Engineering Analysis: Interactive Methods and Programs with FORTRAN, QuickBASIC, MATLAB, and Mathematica

Acquiring Editor: Project Editor: Cover design:

Cindy Renee Carelli Albert W. Starkweather, Jr. Dawn Boyd

Library of Congress Cataloging-in-Publication Data Catalog record is available from the Library of Congress This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage or retrieval system, without prior permission in writing from the publisher. The consent of CRC Press LLC does not extend to copying for general distribution, for promotion, for creating new works, or for resale. Specific permission must be obtained in writing from CRC Press LLC for such copying. Direct all inquiries to CRC Press LLC, 2000 Corporate Blvd., N.W., Boca Raton, Florida 33431. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are only used for identification and explanation, without intent to infringe. Mathematica® is developed by Wolfram Research, Inc., Champaign, IL. Windows® is developed by Microsoft Corp., Redmond, WA.

© 1999 by CRC Press LLC No claim to original U.S. Government works International Standard Book Number 0-8493-2016-X Printed in the United States of America 1 2 3 4 5 6 7 8 9 0 Printed on acid-free paper © 2001 by CRC Press LLC

2. has the following files: Select.BAS Bairstow.EXE EigenStb.EXE ExactFit.EXE LeastSq1.EXE OdeBvpFD.EXE Relaxatn.EXE Bairstow.QB EigenStb.QB ExactFit.QB LagrangI.QB MatxMtpy.QB OdeBvpRK.QB Volume.QB

Select.EXE BRUN40.EXE EigenVec.EXE FindRoot.EXE LeastSqG.EXE OdeBvpRK.EXE RungeKut.EXE CharacEq.QB EigenVec.QB FindRoot.QB LeastSq1.QB NewRaphG.QB ParabPDE.QB WavePDE.QB

CharacEq.EXE EigenVib.EXE Gauss.EXE MatxInvD.EXE ParabPDE.EXE Volume.EXE CubeSpln.QB EigenVib.QB GauJor.QB LeastSqG.QB NuIntgra.QB Relaxatn.QB

CubeSpln.EXE EigenvIt.EXE LagrangI.EXE NuIntgra.EXE QB.EXE DiffTabl.QB EigenvIt.QB Gauss.QB MatxAlgb.QB OdeBvpFD.QB RungeKut.QB

To commence QuickBASIC, when a:\ is prompted on screen, the user enters QB. QB.EXE and BRUN40.EXE therefore are included in . The program Select enables user to select the available QuickBASIC program in this textbook. After user responds with C:\Select, the screen shows a menu as shown in Figure 1 and user then follow the screen help-messages to run a desired program. 3. <mFiles> is a subdirectory associated with MATLAB and has the following files: BVPF.m F.m FuncZ.m integrnd.m ParabPDE.m WavePDE.m

DerivatF.m FindRoot.m FuncZnew.m LagrangI.m Relaxatn.m

DiffTabl.m FP.m FunF.m LeastSqG.m Volume.m

EigenvIt.m Functns.m GauJor.m NewRaphG.m Warping.m

When the 3.5-inch disk containing all of these m files is in Drive A, any of these files can be accessed by enclosing the filename inside a pair of parentheses as illustrated in Section 3.2 where F.m and FP.m are required for FindRoot.m and in Section 5.2 where an integrand function integrnd.m is defined for numerical integration. If all files have been added into MATLAB library m files, then no reference to the Drive A is necessary and the pair of parentheses can also be dropped. 4. <Mathtica> is a subdirectory associated with Mathematica and has the files of: Bairstow.MTK ExactFit.MTK Gauss.MTK LeastSqG.MTK OdeBvpFD.MTK RungeKut.MTK

© 2001 by CRC Press LLC

CubeSpln.MTK FindRoot.MTK GauJor.MTK MatxAlgb.MTK OdeBvpRK.MTK Volume.MTK

DiffTabl.MTK FUNCTNS.MTK LagrangI.MTK NewRaphG.MTK ParabPDE.MTK WavePDE.MTK

EigenVec.MTK EigenvIt.MTK LeastSq1.MTK NuIntgra.MTK Rexalatn.MTK

thank my long-time friends Dr. H. C. Wang, formerly with the IBM Thomas Watson Research Laboratory and now with the Industrial Research Institutes in Hsingchu, Taiwan; Dr. Erik L. Ritman of the Mayo Clinic in Rochester, MN, and Leon Hill of the Boeing Company in Seattle, WA, for their help and encouragement throughout my career in the CAE field. Profs. R. T. DeLorm, L. Kersten, C. W. Martin, R. N. McDougal, G. M. Smith, and E. J. Marmo had assisted in acquiring equipment and research funds which made my development in the CAE field possible, I extend my most sincere gratitude to these colleagues at the University of Nebraska–Lincoln. For providing constructive inputs to my published works, I should give credits to Prof. Gary L. Kinzel of the Ohio State University, Prof. Donald R. Riley of the University of Minnesota, Dr. L. C. Chang of the General Motors’ EDS Division, Dr. M. Maheshiwari and Mr. Steve Zitek of the Brunswick Corp., my former graduate assistants J. Nikkola, T. A. Huang, K. A. Peterson, Dr. W. T. Kao, Dr. David S. S. Shy, C. M. Lin, R. M. Sedlacek, L. Shi, J. D. Wilson, Dr. A. J. Wang, Dave Breiner, Q. W. Dong, and Michael Newman, and former students Jeff D. Geiger, Tim Carrizales, Krishna Pendyala, S. Ravikoti, and Mark Smith. I should also express my appreciation to the readers of my other four textbooks mentioned above who have frequently contacted me and provided input regarding various topics that they would like to be considered as connected to the field of CAE and numerical problems that they would like to be solved by application of computer. Such input has proven to be invaluable to me in preparation of this text. CRC Press has been a delightful partner in publishing my previous book and again this book. The completion of this book would not be possible without the diligent effort and superb coordination of Cindy Renee Carelli, Suzanne Lassandro, and Albert Starkweather, I wish to express my deepest appreciation to them and to the other CRC editorial members. Last but not least, I thank my wife, Rosaline, for her patience and encouragement. Y. C. Pao

© 2001 by CRC Press LLC

5.3 Program Volume — Numerical Solution of Double Integral 5.4 Problems 5.5 References 6 Ordinary Differential Equations — Initial and Boundary Value Problems 6.1 Introduction 6.2 Program RungeKut — Application of Runge-Kutta Method for Solving InitialValue Problems 6.3 Program OdeBvpRK — Application of Runge-Kutta Method for Solving Boundary Value Problems 6.4 Program OdeBvpFD — Application of Finite-Difference Method for Solving Boundary-Value Problems 6.5 Problems 6.6 References 7 Eigenvalue and Eigenvector Problems 7.1 Introduction 7.2 Programs EigenODE.Stb and EigenODE.Vib — for Solving Stability and Vibration Problems 7.3 Program CharacEq — Derivation of Characteristic Equation of a Specified Square Matrix 7.4 Program EigenVec — Solving Eigenvector by Gaussian Elimination Method 7.5 Program EigenvIt — Iterative Solution of Eigenvalue and Eigenvector 7.6 Problems 7.7 References 8 Partial Differential Equations 8.1 Introduction 8.2 Program ParabPDE — Numerical Solution of Parabolic Partial Differential Equations 8.3 Program Relaxatn — Solving Elliptical Partial Differential Equations by Relaxation Method 8.4 Program WavePDE — Numerical Solution of Wave Problems Governed by Hyperbolic Partial Differential Equations 8.5 Problems 8.6 References

© 2001 by CRC Press LLC

1

Matrix Algebra and Solution of Matrix Equations

1.1 INTRODUCTION Computers are best suited for repetitive calculations and for organizing data into specialized forms. In this chapter, we review the matrix and vector notation and their manipulations and applications. Vector is a one-dimensional array of numbers and/or characters arranged as a single column. The number of rows is called the order of that vector. Matrix is an extension of vector when a set of numbers and/or characters are arranged in rectangular form. If it has M rows and N column, this matrix then is said to be of order M by N. When M = N, then we say this square matrix is of order N (or M). It is obvious that vector is a special case of matrix when there is only one column. Consequently, a vector is referred to as a column matrix as opposed to the row matrix which has only one row. Braces are conventionally used to indicate a vector such as {V} and brackets are for a matrix such as [M]. In writing a computer program, DIMENSION or DIM statements are necessary to declare that a certain variable is a vector or a matrix. Such statements instruct the computer to assign multiple memory spaces for keeping the values of that vector or matrix. When we deal with a large number of different entities in a group, it is better to arrange these entities in vector or matrix form and refer to a particular entity by specifying where it is located in that group by pointing to the row (and column) number(s). Such as in the case of having 100 numbers represented by the variable names A, B, …, or by A(1) through A(100), the former requires 100 different characters or combinations of characters and the latter certainly has the advantage of having only one name. The A(1) through A(100) arrangement is to adopt a vector; these numbers can also be arranged in a matrix of 10 rows and 10 columns, or 20 rows and five columns depending on the characteristics of these numbers. In the cases of collecting the engineering data from tests of 20 samples during five different days, then arranging these 100 data into a matrix of 20 rows and five columns will be better than of 10 rows and 10 columns because each column contains the data collected during a particular day. In the ensuing sections, we shall introduce more definitions related to vector and matrix such as transpose, inverse, and determinant, and discuss their manipulations such as addition, subtraction, and multiplication, leading to the organizing of systems of linear algebraic equations into matrix equations and to the methods of finding their solutions, specifically the Gaussian Elimination method. An apparent application of the matrix equation is the transformation of the coordinate axes by a © 2001 by CRC Press LLC

rotation about any one of the three axes. It leads to the derivation of the three basic transformation matrices and will be elaborated in detail. Since the interactive operations of modern personal computers are emphasized in this textbook, how a simple three-dimensional brick can be displayed will be discussed. As an extended application of the display monitor, the transformation of coordinate axes will be applied to demonstrate how animation can be designed to simulate the continuous rotation of the three-dimensional brick. In fact, any threedimensional object could be selected and its motion animated on a display screen. Programming languages, FORTRAN, QuickBASIC, MATLAB, and Mathematica are to be initiated in this chapter and continuously expanded into higher levels of sophistication in the later chapters to guide the readers into building a collection of their own programs while learning the computational methods for solving engineering problems.

1.2 MANIPULATION OF MATRICES Two matrices [A] and [B] can be added or subtracted if they are of same order, say M by N which means both having M rows and N columns. If the sum and difference matrices are denoted as [S] and [D], respectively, and they are related to [A] and [B] by the formulas [S] = [A] + [B] and [D] = [A]-[B], and if we denote the elements in [A], [B], [D], and [S] as aij, bij, dij, and sij for i = 1 to M and j = 1 to N, respectively, then the elements in [S] and [D] are to be calculated with the equations: sij = a ij + b ij

(1)

d ij = a ij − b ij

(2)

and

Equations 1 and 2 indicate that the element in the ith row and jth column of [S] is the sum of the elements at the same location in [A] and [B], and the one in [D] is to be calculated by subtracting the one in [B] from that in [A] at the same location. To obtain all elements in the sum matrix [S] and the difference matrix [D], the index i runs from 1 to M and the index j runs from 1 to N. In the case of vector addition and subtraction, only one column is involved (N = 1). As an example of addition and subtraction of two vectors, consider the two vectors in a two-dimensional space as shown in Figure 1, one vector {V1} is directed from the origin of the x-y coordinate axes, point O, to the point 1 on the x-axis which has coordinates (x1,y1) = (4,0) and the other vector {V2} is directed from the origin O to the point 2 on the y-axis which has coordinates (x2,y2) = (0,3). One may want to find the resultant of {R} = {V1} + {V2} which is the vector directed from the origin to the point 3 whose coordinates are (x3,y3) = (4,3), or, one may want to find the difference vector {D} = {V1} – {V2} which is the vector directed from the origin O to the point 4 whose coordinates are (x4,y4) = (4,–3). In fact, the vector {D} can be obtained by adding {V1} to the negative image of {V2}, namely {V2–} which is a vector directed from the origin O to the point 5 whose coordinates are (x5,y5). Mathematically, based on Equations 1 and 2, we can have: © 2001 by CRC Press LLC

and {D} shown in Figure 1, the calculation of their lengths would need to know the components of these vectors in the coordinate axes and then apply the Pythagorean theorem. Since the vector {R} has components equal to rx = 4 and ry = 3 units along the x- and y-axis, respectively, its length, here denoted with the symbol  , is:

{R} = [rx2 + ry2 ] = [42 + 32 ] = 5 0.5

0.5

(3)

To facilitate the calculation of the length of a generalized vector {V} which has N components, denoted as v1 through vN, its length is to be calculated with the following formula obtained from extending Equation 3 from two-dimensions to Ndimensions:

{V} = [v12 + v22 + … + v2N ]

0.5

(4)

For example, a three-dimensional vector has components v1 = vx = 4, v2 = vy = 3, and v3 = vz = 12, then the length of this vector is {V} = [42 + 32 + 122]0.5 = 13. We shall next show that Equation 4 can also be derived through the introduction of the multiplication rule and transposition of matrices.

1.2 MULTIPLICATION OF MATRICES A matrix [A] of order L (rows) by M (columns) and a matrix [B] of order M by N can be multiplied in the order of [A][B] to produce a new matrix [P] of order L by N. [A][B] is said as [A] post-multiplied by [B], or, [B] pre-multiplied by [A]. The elements in [P] denoted as pij for i = 1 to N and j = 1 to M are to be calculated by the formula: p ij =

∑

M k =1

a ik b kj

(5)

Equation 5 indicates that the value of the element pij in the ith row and jth column of the product matrix [P] is to be calculated by multiplying the elements in the ith row of the matrix [A] by the corresponding elements in the jth column of the matrix [B]. It is therefore evident that the number of elements in the ith row of [A] should be equal to the number of elements in the jth column of [B]. In other words, to apply Equation 5 for producing a product matrix [P] by multiplying a matrix [A] on the right by a matrix [B] (or, to say multiplying a matrix [B] on the left by a matrix [A]), the number of columns of [A] should be equal to the number of row of [B]. A matrix [A] of order L by M can therefore be post-multiplied by a matrix [B] of order M by N; but [A] cannot be pre-multiplied by [B] unless L is equal to N! As a numerical example, consider the case of a square, 3 × 3 matrix postmultiplied by a rectangular matrix of order 3 by 2. Since L = 3, M = 3, and N = 2, the product matrix is thus of order 3 by 2. © 2001 by CRC Press LLC

1 4   7

3  6 6 5 9 4

2 5 8

−3 1(6) + 2(5) + 3( 4) 1( −3) + 2( −2) + 3( −1)    −2  = 4(6) + 5(5) + 6( 4) 4( −3) + 5( −2) + 6( −1) −1  7(6) + 8(5) + 9( 4) 7( −3) + 8( −2) + 9( −1)  6 + 10 + 12 =  24 + 25 + 24 42 + 40 + 32

−3 − 4 − 3   28 −12 − 10 − 5 =  73 −21 − 16 − 9  114

−10  −27 −46

More exercises are given in the Problems listed at the end of this chapter for the readers to practice on the matrix multiplications based on Equation 5. It is of interest to note that the square of the length of a vector {V} which has N components as defined in Equation 4, {V}2, can be obtained by application of Equation 5 to {V} and its transpose denoted as {V}T which is a row matrix of order 1 by N (one row and N columns). That is:

{V} = {V}T {V} = v11 + v22 + … + v32 2

(6)

For a L-by-M matrix having elements eij where the row index i ranges from 1 to L and the column index j ranges from 1 to M, the transpose of this matrix when its elements are designated as trc will have a value equal to ecr where the row index r ranges from 1 to M and the column index c ranges from 1 to M because this transpose matrix is of order M by L. As a numerical example, here is a pair of a 3 × 2 matrix [G] and its 2 × 3 transpose [H]: 6 [G] = 5 3× 2 4

−3 −2  and −1

6 −3

[H] = [G]T =  2 ×3

5 −2

4 −1

If the elements of [G] and [H] are designated respectively as gij and hij, then hij = gji. For example, from above, we observe that h12 = g21 = 5, h23 = g32 = –1, and so on. There will be more examples of applications of Equations 5 and 6 in the ensuing sections and chapters. Having introduced the transpose of a matrix, we can now conveniently revisit the addition of {D} and {E} in Figure 1 in algebraic form as {F} = {D} + {E} = [4 –3]T + [–3 1]T = [4+(–3) –3+1]T = [1 –2]T. The resulting sum vector is indeed correct as it is graphically verified in Figure 1. The saving of space by use of transposes of vectors (row matrices) is not evident in this case because all vectors are two-dimensional; imagine if the vectors are of much higher order. Another noteworthy application of matrix multiplication and transposition is to reduce a system of linear algebraic equations into a simple, (or, should we say a single) matrix equation. For example, if we have three unknowns x, y, and z which are to be solved from the following three linear algebraic equations: © 2001 by CRC Press LLC

x + 2 y + 3z = 4 5x + 6y + 7z = 8 −2 x − 37 = 9

(7)

Let us introduce two vectors, {V} and {R}, which contain the unknown x, y, and z, and the right-hand-side constants in the above three equations, respectively. That is: x  T V = x y z = { } [ ] y  z 

and

4 T R = 4 8 9 = { } [ ] 8 9 

(8)

Then, making use of the multiplication rule of matrices, Equation 5, the system of linear algebraic equations, 7, now can be written simply as:

[C]{V} = {R}

(9)

where the coefficient matrix [C] formed by listing the coefficients of x, y, and z in first equation in the first row and second equation in the second row and so on. That is, 1 [C] =  5 −2

2 6 −3

3 7 0 

There will be more applications of matrix multiplication and transposition in the ensuing chapters when we discuss how matrix equations, such as [C]{V} = {R}, can be solved by employing the Gaussian Elimination method, and how ordinary differential equations are approximated by finite differences will lead to the matrix equations. In the abbreviated matrix form, derivation and explanation of computational methods becomes much simpler. Also, it can be observed from the expressions in Equation 8 how the transposition can be conveniently used to define the two vectors not using the column matrices which take more lines.

FORTRAN VERSION Since Equations 1 and 2 require repetitive computation of the elements in the sum matrix [S] and difference matrix [D], machine could certainly help to carry out this laborous task particularly when matrices of very high order are involved. For covering all rows and columns of [S] and [D], looping or application of DO statement of the FORTRAN programming immediately come to mind. The following program is provided to serve as a first example for generating [S] and [D] of two given matrices [A] and[B]: © 2001 by CRC Press LLC

A= 1

2

3

4

Notice that the elements of [A] should be entered row by row. While the rows are separated by ;, in each row elements are separated by comma. After the print out of the above results, >> sign will again appear. To eliminate the unnecessary line space (between A = and the first row 1 2), the statement format compact can be entered as follows (the phrase “pressing ENTER key” will be omitted from now on): >> format compact, B = [5,6;7,8] B= 5

6

7

8

Notice that comma is used to separate the statements. To demonstrate matrix subtraction and addition, we can have: >> A-B ans = –4

–4

–4

–4

>> A + B ans = 6

8

10

12

To apply MATLAB for transposition and multiplication of matrices, we can have: >> C = [1,2,3;4,5,6] C= 1

2

3

4

5

6

>> C' ans = 1

4

2

5

3

6

© 2001 by CRC Press LLC

>> D = [1,2,3;4,5,6]; E = [1,2;2,3;3,4]; P = D*E P= 14

20

32

47

Notice that MATLAB uses ' (single quote) in place of the superscripted symbol T for transposition. When ; (semi-colon) follows a statement such as the D statement, the results will not be displayed. As in FORTRAN and QuickBASIC, * is the multiplication operator as is used in P = D*E, here involving three matrices not three single variables. More examples of MATLAB applications including plotting will ensue. To terminate the MATLAB operation, simply enter quit and then the RETURN key.

MATHEMATICA APPLICATIONS To commence the service of Mathematica from Windows setup, simply point the mouse to it and double click the left button. The Input display bar will appear on screen, applications of Mathematica can start by entering commands from keyboard and then press the Shift and Enter keys. To terminate the Mathematica application, enter Quit[] from keyboard and then press the Shift and Enter keys. Mathematica commands, statements, and functions are gradually introduced and applied in increasing degree of difficulty. Its graphic capabilities are also utilized in presentation of the computed results. For matrix operations, Mathematica can compute the sum and difference of two matrices of same order in symbolic forms, such as in the following cases of involving two matrices, A and B, both of order 2 by 2: In[1]: = A = {{1,2},{3,4}}; MatrixForm[A] Out[1]//MatrixForm = 1

2

3

4

In[1]: = is shown on screen by Mathematica while user types in A = {{1,2},{3,4}}; MatrixForm[A]. Notice that braces are used to enclose the elements in each row of a matrix, the elments in a same row are separated by commas, and the rows are also separated by commas. MatrixForm demands that the matrix be printed in a matrix form. Out[1]//MatrixForm = and the rest are response of Mathematica. In[2]: = B = {{5,6},{7,8}}; MatrixForm[B] Out[2]//MatrixForm = 5 7

6 8

© 2001 by CRC Press LLC

In[3]: = MatrixForm[A + B] Out[3]//MatrixForm = 6

8

10

12

In[4]: = Dif = A-B; MatrixForm[Dif] Out[4]//MatrixForm = –4

–4

–4

–4

In[3] and In[4] illustrate how matrices are to be added and subtracted, respectively. Notice that one can either use A + B directly, or, create a variable Dif to handle the sum and difference matrices. Also, Mathematica has a function called Transpose for transposition of a matrix. Let us reuse the matrix A to demonstrate its application: In[5]: = AT = Transpose[A]; MatrixForm[AT] Out[5]//MatrixForm = 1

3

2

4

1.3 SOLUTION OF MATRIX EQUATION Matrix notation offers the convenience of organizing mathematical expression in an orderly manner and in a form which can be directly followed in coding it into programming languages, particularly in the case of repetitive computation involving the looping process. The most notable situation is in the case of solving a system of linear algebraic equation. For example, if we need to determine a linear equation y = a1 + a2x which geometrically represents a straight line and it is required to pass through two specified points (x1,y1) and (x2,y2). To find the values of the coefficients a1 and a2 in the y equation, two equations can be obtained by substituting the two given points as:

(1)a1 + (x1 )a 2 = y1

(1)

(1)a1 + (x 2 )a 2 = y2

(2)

and

© 2001 by CRC Press LLC

1.4 PROGRAM GAUSS Program Gauss is designed for solving N unknowns from N simultaneous, linear algebraic equations by the Gaussian Elimination method. In matrix notation, the problem can be described as to solve a vector {X} from the matrix equation:

[C]{X} = {V}

(1)

where [C] is an NxN coefficient matrix and {V} is a Nx1 constant vector, and both are prescribed. For example, let us consider the following system: 9x1 + x 2 + x3 = 10

(2)

3x1 + 6x 2 + x3 = 14

(3)

2 x1 + 2 x 2 + 3x3 = 3

(4)

If the above three equations are expressed in matrix form as Equation 1, then: 9 [C] = 3 2

1 6 2

1 1, 3

10  {V} = 14,  3 

(5,6)

and  x1  X = { } x 2  = x1 x 2 x3  x3 

[

]

T

(7)

where T designates the transpose of a matrix.

GAUSSIAN ELIMINATION METHOD A systematic procedure named after Gauss can be employed for solving x1, x2, and x3 from the above equations. It consists of first dividing Equation 28 by the leading coefficient, 9, to obtain: x1 +

1 1 10 x + x = 9 2 9 3 9

(8)

This step is called normalization of the first equation of the system (1). The next step is to eliminate x1 term from the other (in this case, two) equations. To do that, we multiply Equation 8 by the coefficients associated with x1 in Equations 3 and 4, respectively, to obtain: © 2001 by CRC Press LLC

to solve for {X} when the coefficient matrix [A] and the right-hand side vector {R} are both provided. The following is an example of interactive application: In[1]: = A = {{3,6,14},{6,14,36},{14,36,98}} Out[1]: = {{3, 6, 14}, {6, 14, 36}, {14, 36, 98}} In[2]: = MatrixForm[A] Out[2]//MatrixForm: = 3

6

14

6

14

36

14

36

98

In[3]: = R = {9,20,48} Out[3]: = {9, 20, 48} In[4]: = LinearSolve[A,R] Out[4]: = {–9,13,–3} Output[2] and Output[1] demonstrate the difference in display of matrix [A] when MatrixeForm is requested, or, not requested, respectively. It shows that without requesting of MatrixForm, some screen space saving can be gained. Output[4] gives the solution {X} = [–9 13 –3]T for the matrix equation [A]{X} = {R} where the coefficient matix [A] and vector {R} are provided by Input[1] and Input[3], respectively.

1.5 MATRIX INVERSION, DETERMINANT, AND PROGRAM MatxInvD Given a square matrix [C] of order N, its inverse as [C]–1 of the same order is defined by the equation:

[C][C]−1 = [C]−1[C] = [I]

(1)

where [I] is an identity matrix having elements equal to one along its main diagonal and equal to zero elsewhere. That is: . 1 0 0 1 0 [I] =   . . .  . 0 0 © 2001 by CRC Press LLC

. . . .

. . . .

. . . .

0 0    1

(2)

MATHEMATICA APPLICATIONS Mathematica has a function called Inverse for inversion of a matrix. Let us reuse the matrix A that we have entered in earlier examples and continue to demonstrate the application of Inverse: In[1]: = A = {{1,2},{3,4}}; MatrixForm[A] Out[1]//MatrixForm = 1

2

3

4

In[2]: = B = {{5,6},{7,8}}; MatrixForm[B] Out[2]//MatrixForm = 5

6

7

8

In[3]: = MatrixForm[A + B] Out[3]//MatrixForm = 6

8

10

12

In[4]: = Dif = A-B; MatrixForm[Dif] Out[4]//MatrixForm = –4

–4

–4

–4

In[5]: = AT = Transpose[A]; MatrixForm[AT] Out[5]//MatrixForm = 1

3

2

4

In[6]: = Ainv = Inverse[A]; MatrixForm[Ainv] Out[6]//MatrixForm = –2

1

3  1 − 2  2

© 2001 by CRC Press LLC

To verify whether or not the inverse matrix Ainv obtained in Output[6] indeed satisfies the equations [A][A]–1 = [I] which is the identity matrix, we apply Mathematica for matrix multiplication: In[7]: = Iden = A.Ainv; MatrixForm[Iden] Out[7]//MatrixForm = 1

0

0

1

A dot is to separate the two matrices A and Ainv which is to be multiplied in that order. Output[7] proves that the computed matrix, Ainv, is the inverse of A! It should be noted that D and I are two reserved variables in Mathematica for the determinant of a matrix and the identity matrix. In their places, here Dif and Iden are adopted, respectively. For further testing, we show that [A][A]T is a symmetric matrix: In[8]: = S = A.AT; MatrixForm[S] Out[8]//MatrixForm = 5

11

11

25

And, the unknown vector {X} in the matrix equation [A]{X} = {R} can be solved easily if {R} is given and [A]–1 are available: In[9]: = R = {13,31}; X = Ainv.R Out[9] = {5, 4} The solution of x1 = 5 and x2 = 4 do satisfy the equations x1 + 2x2 = 13 and 3x1 + 4x2 = 31.

TRANSFORMATION

OF

COORDINATE SYSTEMS, ROTATION,

AND

ANIMATION

Matrix algebra can be effectively applied for transformation of coordinate systems. When the cartesian coordinate system, x-y-z, is rotated by an angle z about the z-axis to arrive at the system x-y-z as shown in Figure 2, where z and z axes coincide and directed outward normal to the plane of paper, the new coordinates of a typical point P whose coordinates are (xP,yP,zP) can be easily obtained as follows: x ′P = OP cos(θ P − θ z ) = (OP cos θ P ) cos θ z + (OP sin θ P ) sin θ z = x P cos θ z + y p sin θ z y ′P = OP sin(θ P − θ z ) = (OP sin θ P ) cos θ z − (OP cos θ P ) sin θ z = x p sin θ z + y p sin θ z © 2001 by CRC Press LLC

4. Apply the QuickBASIC and FORTRAN versions of the program MatxAlgb to verify the results of Problems 1, 2, and 3. 5. Repeat Problem 4 but use MATLAB. 6. Apply the program MatxInvD to find [C]–1 of the matrix [C] given in Problem 1 and also to ([C]T)–1. Is ([C]–1)T equal to ([C]T)–1? 7. Repeat Problem 6 but use MATLAB. 8. For statistical analysis of a set of N given data X1, X2, …, XN, it is often necessary to calculate the mean, m, and standard deviation, 5, by use of the formulas: m=

1 (X + X2 + … + X N ) N 1

and

[

]

2 2 2 1 σ =  ( X1 − m ) + ( X 2 − m ) + … + ( X N − m )  N 

0.5

Use indicial notation to express the above two equations and then develop a subroutine meanSD(X,N,RM,SD) for taking the N values of X to compute the real value of mean, RM, and standard deviation, SD. 9. Express the ith term in the following series in indicial notation and then write an interactive program SinePgrm allowing input of the x value to calculate sin(x) by terminating the series when additional term contributes less than 0.001% of the partial sum of series in magnitude: Sin x =

x1 x3 x 5 − + −… 1! 3! 5!

Notice that Sin(x) is an odd function so the series contains only terms of odd powers of x and the series carries alternating signs. Compare the result of the program SinePgrm with those obtained by application of the library function Sin available in FORTRAN and QuickBASIC. 10. Same as Problem 9, but for the cosine series: Cos x = 1 −

x2 x 4 x6 + − +… 2! 4! 6!

Notice that Cos(x) is an even function so the series contains only terms of even powers of x and the series also carries alternating signs. 11. Repeat Problem 4 but use Mathematica. 12. Repeat Problem 6 but use Mathematica. © 2001 by CRC Press LLC

GAUSS 1. Run the program GAUSS to solve the problem: 1 4   7

2 5 8

3   x1   2    6  x 2  =  8  10   x3  14

2. Run the program GAUSS to solve the problem: 0 4   7

2 5 8

3  x1  −1   6 x 2  =  8  9  x3  14 

What kind of problem do you encounter? “Divided by zero” is the message! This happens because the coefficient associated with x1 in the first equation is equal to zero and the normalization in the program GAUSS cannot be implemented. In this case, the order of the given equations needs to be interchanged. That is to put the second equation on top or find below the first equation an equation which has a coefficient associated with x1 not equal to zero and is to be interchanged with the first equation. This procedure is called “pivoting.” Subroutine GauJor has such a feature incorporated, apply it for solving the given matrix equation. 3. Modify the program GAUSS by following the Gauss-Jordan elimination procedure and excluding the back-substitution steps. Name this new program GauJor and test it by solving the matrix equations given in Problems 1 and 2. 4. Show all details of the normalization, elimination, and backward substitution steps involved in solving the following equations by application of Gaussian Elimination method: 4x1 + 2x2 – 3x3 = 8 5x1 – 3x2 + 7x3 = 26 –x1 + 9x2 – 8x3 = –10 5. Present every normalization and elimination steps involved in solving the following system of linear algebraic equations by the Gaussian Elimination Method: 5x1 – 2x2 + 2x3 = 9, –2x1 + 7x2 – 2x3 = 9, and 2x1 – 2x2 + 9x3 = 41

© 2001 by CRC Press LLC

6. Apply the Gauss-Jordan elimination method to solve for x1, x2, and x3 from the following equations: 0 2  4

−1  x1  1   3  x 2  = 1 7   x3  1

1 9 24

Show every normalization, elimination, and pivoting (if necessary) steps of your calculation. 7. Solve the matrix equation [A]{X} = {C} by Gauss-Jordan method where: 3 2  4

2 5 1

1   x1  −2    −1 x 2  =  −3 7   x3   3 

Show every interchange of rows (if you are required to do pivoting before normalization), normalization, and elimination steps by indicating the changes in [A] and {C}. 8. Apply the program GauJor to solve Problem 7. 9. Present every normalization and elimination steps involved in solving the following system of linear algebraic equations by the Gauss-Jordan Elimination Method: 5x1 – 2x2 + x3 = 4 –2x1 + 7x2 – 2x3 = 9 x1 – 2x2 + 9x3 = 40 10. 11. 12. 13. 14.

Apply the program Gauss to solve Problem 9 described above. Use MATLAB to solve the matrix equation given in Problem 7. Use MATLAB to solve the matrix equation given in Problem 9. Use Mathematica to solve the matrix equation given in Problem 7. Use Mathematica to solve the matrix equation given in Problem 9.

MATRIX INVERSION 1. Run the program MatxInvD for finding the inverse of the matrix: 3 [A] = 0 2 © 2001 by CRC Press LLC

0 5 0

2 0  3

2. Write a program Invert3 which inverts a given 3 × 3 matrix [A] by using the cofactor method. A subroutine COFAC should be developed for calculating the cofactor of the element at Ith row and Jth column of [A] in term of the elements of [A] and the user-specified values of I and J. Let the inverse of [A] be designated as [AI] and the determinant of [A] be designated as D. Apply the developed program Invert3 to generate all elements of [AI] by calling the subroutine COFAC and by using D. 3. Write a QuickBASIC or FORTRAN program MatxSorD which will perform the addition and subtraction of two matrices of same order. 4. Write a QuickBASIC or FORTRAN program MxTransp which will perform the transposition of a given matrix. 5. Translate the FORTRAN subroutine MatxMtpy into a MATLAB m file so that by entering the matrices [A] and [B] of order L by M and M by N, respectively, it will produce a product matrix [P] of order L by N. 6. Enter MATLAB commands interactively first a square matrix [A] and then calculate its trace. 7. Use MATLAB commands to first define the elements in its upper right corner including the diagonal, and then use the symmetric properties to define those in the lower left corner. 8. Convert either QuickBasic or FORTRAN version of the program MatxInvD into a MATLAB function file MatxInvD.m with a leading statement function [Cinv,D] = MatxInvD(C,N) 9. Apply the program MatxInvD to invert the matrix: 1 [A] = 5 8

3 6 9

4 7  10 

Verify the answer by using Equation 1. 10. Repeat Problem 9 but by MATLAB operation. 11. Apply the program MatxInvD to invert the matrix: −9 [A] =  −3 −6

−1 −4 −7

−2  −5 −8

Verify the answer by using Equation 1. 12. Repeat Problem 11 but by MATLAB operations. 13. Derive [Rx] and verify that it is indeed equal to [Tx]T. Repeat for [Ry] and [Rz]. 14. Apply MATLAB to generate a matrix [Rz] for θz = 45° and then to use [Rz] to find the rotated coordinates of a point P whose coordinates before rotation are (1,–2,5). © 2001 by CRC Press LLC

2

Exact, Least-Squares, and Cubic Spline Curve-Fits

2.1 INTRODUCTION Engineers conduct experiments and collect data in the laboratories. To make use of the collected data, these data often need to be fitted with some particularly selected curves. For example, one may want to find a parabolic equation y = c1 + c2x + c3x2 which passes three given points (xi,yi) for i = 1,2,3. This is a problem of exact curvefit. Or, since knowing in advance that these three points should all fall on a straight line, but the reason that they are not is because of bad calibration of the measuring equipment or because of presence of noises in the testing environment. In case that we may want express this straight line by the equation y = c1 + c2x for the stress and strain data collected for a stretching test of a metal bar in the elastic range, then the question of how to determine the two coefficients c1 and c2 is a matter of deciding on which criterion to adopt. The Least-Squares method is one of the criteria which is most popularly used. The two cases cited are the consideration of adopting the two and three lowest polynomial terms, x0, x1, and x2, and linearly combining them. If the collected data are supposed to represent a sinusoidal function of time, the curve to be determined may have to be assumed as x(t) = c1sint + c2sin3t + c3sin5t + c4sin7t by linearly combining 4 odd sine terms. This is the case of selecting four particular functions, namely, fi(t) = sin(2i–1)t for i = 1,2,3,4., and to determine the coefficients c1–4 by application of the least-squares method. Often some special form of curve needs to be selected to fit a given set of data, the least-squares criterion can still be applied if mathematical transformations can be found to convert the equation describing the curve into linear equations. This is discussed in a section devoted to transformed least-squares curve-fit. Another commonly applied curve-fit technique is the cubic spline method which allows smooth cubic equations to be derived to ensure continuous slopes and curvatures passing all given points. The mathematics involved in this method will be presented. In the following sections, we shall discuss the development of the programs ExactFit, LeastSq1, LeastSqG, and CubeSpln for the four curve-fit needs mentioned above.

2.2 EXACT CURVE FIT As another example of solving a matrix equation, let us consider the problem of finding a parabolic equation y = c1 + c2x + c3x2 which passes three given points © 2001 by CRC Press LLC

the coefficients c1 and c2 can be uniquely computed. If these three points are not all on a straight line and they still need to be fitted by a linear equation Y = c1 + c2X, then they are forced to be fitted by a particular straight line based on a selected criterion, permitting errors to exist between the data points and the line. The least-squares curve-fit for finding a linear equation Y = c1 + c2X best representing the set of N given points, denoted as (Xi,Yi) for i = 1 to N, is to minimize the errors between the straight line and the points to be as small as possible. Since the given points may be above, on, or below the straight line, and these errors may cancel from each other if they are added together, we consider the sum of the squares of these errors. Let us denote yi to be the value of Y at Xi and S be the sum of the errors denoted as ei for i = 1 to N, then we can write: N

S = e12 + e 22 + … + e 2N =

∑e

2 i

(1)

i =1

where for i = 1,2,…,N e i = y i − y i = c1 + c2 X i − Yi

(2)

It is obvious that since Xi and Yi are constants, the sum S of the errors is a function of ci and c2. To find a particular set of values of c1 and c2 such that S reaches a minimum, we may therefore base on calculus3 and utilize the conditions ∂S/∂c1 = 0 and ∂S/∂c2 = 0. From Equation 1, we have the partial derivatives of S as:  ∂e ∂e  ∂e ∂S = 2 e1 1 + e 2 2 + … + e N N  = 2 ∂c1 ∂ c ∂ c ∂c1   1 1

N

∂e i

∑ e ∂c i

1

i =1

and  ∂e ∂e  ∂e ∂S = 2 e1 1 + e 2 2 + … + e N N  = 2 ∂c2 ∂ c ∂ c ∂c 2   2 2

N

∂e i

∑ e ∂c i

i =1

2

From Equation 2, we note that ∂ei/∂c1 = 1 and ∂ei/∂c2 = Xi. Consequently, the two extremum conditions lead to two algebraic equations for solving the coefficients c1 and c2:    © 2001 by CRC Press LLC

N

∑ i =1

  1 c1 +   

N

∑ i =1

 X i  c2 = 

N

∑Y

i

i =1

(3)

2.4 PROGRAM LEASTSQG — GENERALIZED LEAST-SQUARES CURVEFIT Program LeastSqG is designed for curve-fitting of a set of given data by a linear combination of a set of selected functions based on the least-squares criterion.2 Let us consider N points whose coordinates are (Xk,Yk) for k = 1 to N and let the M selected function be f1(X) to fM(X) and the equation determined by the leastsquares curve-fit be: M

Y( X) = a1f1 ( X) + a 2 f2 ( X) + … + a M fM (X ) =

∑ a f (X) j j

(1)

j=1

The least-squares curve-fit for finding the coefficients c1–M is to minimize the errors between the computed Y values based on Equation 1 and the given Y values at all Xk’s for k = 1 to N. Let us denote yk to be the value of Y calculated at Xk using Equation 1 and S be the sum of the errors denoted as ek for k = 1 to N. Since the yk could either be greater or less than Yk, these errors ek’s may cancel from each other if they are added together. We therefore consider the sum of the squares of these errors and write: N

S = e12 + e 22 + … + e 2N =

∑e

2 k

(2)

k =1

where for k = 1,2,…,N  e k = y k − Yk =   

M

∑ j=1

 a jfj ( X k ) − Yk  

(3)

It is obvious that since Xk and Yk are constants, the sum S of the errors is a function of a1 to M. To find a particular set of values of a1 to M such that S reaches a minimum, we may therefore base on calculus3 and utilize the conditions ∂S/∂ai = 0 for i = 1 to M. From Equation 2, we can have the partial derivatives of S with respect to ai’s, for i = 1 to M, as: N

 ∂e ∂e  ∂e ∂e ∂S = 2 e1 1 + e 2 2 + … + e N N  = 2 ek k ∂a i ∂a i ∂a i  ∂a i  ∂a1 k =1

∑

From Equation 3, we note that ∂ek/∂ai = fi(Xk). Consequently, the M extremum conditions, ∂S/∂ai = 0 for i = 1 to M, lead to M algebraic equations for solving the coefficients a1 to M. That is, for i = 1 to M: M

 N  fi ( X k )fj ( X k )a j =   k =1 

∑∑ j=1

© 2001 by CRC Press LLC

N

∑ f (X )Y i

k =1

k

k

(4)

derived equation. In case that three points are given and the two selected functions are f1(x) = x–1 and f2(x) = x2, then the interactive operation goes as follows: In[2]: = Fit[{{1,2},{3,5},{4,13}}, {x–1, x^2}, x] Out[2]: = –7.05263 (–1 + x) + 2.13158 x Two other examples previously presented in the MATLAB applications can also be considered and the results are: In[3]: = (Fit[{{1,2}, {2,4}, {3,7}, {4,11}, {5,23}, {6,45}}, {x, x^2, Sin[x], Cos[x]}, x]) Out[3]: = –4.75756 x + 2.11159 x2 – 0.986915 Cos[x] + 5.76573 Sin[x] and In[4]: = (Fit[{{1.4, 2.25}, {3.2, 15}, {4.8, 26.25}, {8, 33}, {10, 35}}, {Sin[Pi*x/20], Sin[3*Pi*x/20], Sin[5*Pi*x/20]}, x]) Pi x  3 Pi x  5 Pi x  Out[4]: = 35.9251 Sin  − 1.19261 Sin  − 3.46705 Sin   20   20   20  All of the results obtained here are in agreement with those presented earlier.

2.5 PROGRAM CUBESPLN — CURVE FITTING WITH CUBIC SPLINE If a set of N given points (Xi,Yi) for i = 1, 2,…,N is to be fitted with a polynomial of N–1 degree passing these points exactly, the polynomial curve will have many fluctuations between data points as the number of points, N, increases. A popular method for avoiding such over-fluctuation is to fit every two adjacent points with a cubic equation and to impose continuity conditions at the point shared by two neighboring cubic equations. This is like using a draftsman’s spline attempting to pass through all of the N given points. It is known as cubic-spline curve fit. For a typical interval of X, say Xi to Xi + 1, the cubic equation can be written as: Y = a i X3 + bi X 2 + ci X + d i

(1)

where ai, bi, ci, and di are unknown coefficients. If there are N–1 intervals and each interval is fitted by a cubic equation, the total number of unknown coefficients is equal to 4(N–1). They are to be determined by the following conditions: 1. Each curve passes two specified points. The first and last specified points are used once by the first and N-1st curves, respectively, whereas all interior points are used twice by the two cubic curves meeting there. This gives 2 + 2(N–2), or, 2N–2 conditions. © 2001 by CRC Press LLC

2. Every two adjacent cubic curves should have equal slope (first derivative with respect to X) at the interior point which they share. This gives N–2 conditions. (2) 3. For further smoothness of the curve fit, every two adjacent cubic curves should have equal curvature (second derivative with respect to X) at the interior point which they share. This gives N–2 conditions. 4. At the end points, X1 and XN, the nature spline requires that the curvatures be equal to zero. This gives 2 conditions. Instead of solving the coefficients in Equation 1, the usual approach is to apply the above-listed conditions for finding the curvatures, Y″ at the interior points, that is for i = 2,3,…,N–1 since (Y″)1 = (Y″)N = 0. To do so, we notice that if Y is a cubic polynomial of X, Y″ is then linear in X and can be expressed as: Y′′ = AX + B

(3)

If this is used to fit the ith interval, for which the increment in X is here denoted as Hi = Xi + 1–Xi, we may replace the unknown coefficients A and B with the curvatures at Xi and Xi + 1, (Y″)i and (Y″)i + 1 by solving the two equations: Yi′′ = AX i + B

and

Yi′′+1 = AX i +1 + B

(4,5)

By using the Cramer’s rule, it is easy to obtain: A=

Yi′+1 − Yi′′ X i +1 − X i

and

B = Y′′ −

X i (Yi′′+1 − Yi′′) X i +1 − X i

(6,7)

Consequently, Equation 3 can be written as: Y′′ =

Yi′′ (X − X) + YHi′′+1 (X − Xi ) H i i +1 i

(8)

Equation 8 can be integrated successively to obtain the expressions for Y and Y as: 2 2 − Yi′′ Y′′ (9) Y′ + X i +1 − X) + i +1 ( X − X i ) + C1 ( 2H i 2H i and 3 3 Y′′ Y′′ (10) Y = i ( X i +1 − X) + i +1 ( X − X i ) + C1X + C2 6H i 6H i © 2001 by CRC Press LLC

The integration constants C1 and C2 can be determined by the conditions that at Xi, Y = Yi and at Xi + 1, Y = Yi + 1. Based on Equation 10, the two conditions lead to:

Yi =

Yi′′ 2 H + C1X i + C2 6 i

and

Yi +1 =

Yi′′+1 2 H + C1X i +1 + C2 6 i

(11,12)

Again, Cramer’s rule can be applied to obtain:

C1 =

(Y′′− Y′′ )H i +1

i

6

i

−

Yi − Yi +1 Hi

(13)

and

C2 = −

(Xi+1Yi′′− XiYi′′+1 )Hi − XiYi+1 − Xi+1Yi 6

Hi

Hi

(14)

Substituting C1 and C2 into Equations 11 and 12 and rearranging into terms involving the unknown curvatures, the expressions for the cubic curve are:

Y=

3  Y′′  ( X − X )3  Yi′′  ( X i +1 − X) i  − H i ( X i +1 − X) + i +1  − H i ( X − X i ) Hi 6  6  Hi    

 X − X  X − Xi  + Yi  i +1 + Yi +1     Hi   Hi 

(15)

 H ( X − X )2   ( X − X )2 H  Y − Y i i  + Yi′′+1  Y′ = Yi′′  i − i +1 − i  + i +1 Hi 2H i 6  6   2 H i 

(16)

and

Equation 15 indicates that the cubic curve for the ith interval is completely defined if in addition to the specified values of Yi and Yi + 1, the curvatures at Xi and Xi + 1, (Y″)i and (Y″)i + 1 respectively, also can be found. To find all of the curvatures at the interior points X2 through XN–1, we apply the remaining unused conditions mentioned in (2). That is, matching the slopes of two adjacent cubic curves at these © 2001 by CRC Press LLC

Evaluate calculates the values of the fitted polynomial within the specified interval. In this case, the y values are interpolated using the cubic polynomial. To add more features to a plot, Mathematica has commands Text allowing a string of characters to be placed at a desired location, and FrameLabel allowing labels to be added to the axes. In fact, Figure 8 is not the result of Input[3] but the addition of the five + markers with the statement Input[4]: = Show[%,Graphics[Table[Text[“+”,{x[[i]],y[[i]]}],{i,1,5}]]] Output[4] = —Graphics— % refers to the previous output and %% refers to the next-to-the-last output, and so on. Show and Graphics enable additional graphics to be shown. Table lists a series of entries specified by the running index variable which in this case is I having a starting value of 1 and incremented by 1 (omitted), and ending at a value equal to 5. Notice that the four * markers are not exactly located at the coordinates specified, for the reason that the character itself is not centered but offsets like a superscript. As another example, consider the plot shown in Figure 9 which is resulted from of the following statements: Input[1]: = X = {1,2,3,4} Output[1] = {1, 2, 3, 4} Input[2]: = Y = {2, 4, 7, 13} Output[2] = {2, 4, 7, 13} Input[3]: = X1 = {0.5, 1.2, 2.5, 3.7} Output[3] = {0.5, 1.2, 2.5, 3.7}

© 2001 by CRC Press LLC

Input[4]: = Y1 = {3, 6, 5, 11} Output[4] = {3, 6, 5, 11} Input[5]: = X2 = {3.0, 3.6, 4.2, 5.1} Output[5] = {3.0, 3.6, 4.2, 5.1} Input[6]: = Y2 = {3, 6, 8, 11} Output[6] = {3, 6, 8, 11} Input[7]: = g1 = Show[Graphics[Table[Text[“*”,{X[[i]],Y[[i]] }],{i,1,5}], Table[Text[“ + ”,{X1[[i]],Y1[[i]]}],{i,1,5}], Table[Text[“.”,{X2[[i]],Y2[[i]]}],{i,1,5}]]] Output[7] = —Graphics— Input[8]: = g2 = Show[g1, Frame->True, AspectRatio->1, FrameLabel->{“X”,“Y”}] Output[8] = —Graphics— Input[9]: = Show[g2,Graphics[Text[“X–Y — *”,{0.7,12},{–1,0}], Text[“X1–Y1 — + ”,{0.7,11},{–1,0}], Text[“X2–Y2 — .”,{0.7,10}],{–1,0}]]]

© 2001 by CRC Press LLC

Output[4] = —Graphics— Input[5]: = Show[%,Graphics[Table[Text[“X”,{XYS[[i]]}],{i,1,5}]]] Output[5] = —Graphics— In Input[1], << loads the specified package `SplineFit`. Input[2] illustrates how the coordinates of points can be entered as pairs and ; can be used to suppress the display of the output. “Cubic” spline fit is requested in Input[3] and in Input[5] the ParametricPlot command is applied so that the coordinates of the points on the cubic spline curves are generated using a third parameter. Input[5] also demonstrates how a matrix should be called in a looping situation. Notice that two plots for Output[4] and Output[5] are not presented here, for the reason that Figure 9 already shows all of the needed details. However, it should be mentioned that use of “X” as the character in Text for marking the five XYS points in Input[5] will enable it to be centered at the coordinates specified instead of being upwardly offset as is the case of using the character “*” in Figure 9.

2.6 PROBLEMS EXACT CURVE-FIT 1. Modify the program ExactFit so that the given points (1,3), (3,8), and (4,23) can be exact-fitted by the equation y = c1x + c2x2 + c3x4. 2. Modify the program ExactFit so that the given points (0,0.2), (2,0.5), (5,–0.4), and (7,–0.2) can be exact-fitted by the equation y = c1sinx + c2sin2x + c3sin3x + c4sin4x.

LEASTSQ1 1. Given five points (1,1), (2,3), (3,2), (4,5), and (5,4), calculate the coefficients c1 and c2 in the linear equation y = c1 + c2x which fits the five points by the least-squares method. 2. For a given set of data (1,–2), (2,0), (3,1), and (4,3), two equations have been suggested to fit these points. They are Y = X–2 and Y = (-X2 + 7X–10)/2. Based on the least-squares criterion, which equation should be chosen to provide a better fit? Explain why? 3. During a tensile-strength test of a metallic material the following data (Xi,Yi) for i = 1,2,…,7 where X and Y represent strain (extension per unit length) and stress (force per unit area), respectively, have been collected: X Y

.265 1.03

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.395 1.41

.695 1.71

.955 2.09

1.35 2.42

2.05 2.76

2.45 3.01

(x10–3) (x103)

4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

15. 16.

Fit these data by the least-squares method with an equation Y = b1XExp(b2X) and calculate the values of the coefficients b1 and b2. (Note: ln(Y/X) = ln(b1) + b2X) Apply polyfit of MATLAB to the data given in Problem 1 and fit them linearly. Compare the answer with that of Problem 1. Same as Problem 4 except the data are to be fitted by a quadratic equation. Same as Problem 4 except the data are to be fitted by a cubic equation. Same as Problem 4 except the data are to be fitted by a quartic equation. Use the results of Problems 4 through 7 and enter MATLAB statements to compare their errors. Apply polyfit of MATLAB to the data given in Problem 3 and fit them linearly. Compare the answer with that of Problem 3. Same as Problem 9 except the data are to be fitted by a quadratic equation. Same as Problem 9 except the data are to be fitted by a cubic equation. Same as Problem 9 except the data are to be fitted by a quartic equation. Use the results of Problems 9 through 12 and enter MATLAB statements to compare their errors. Save the results of Problems 4 through 7 and generating the four curves determined by the least-squares method. Obtain a composite graph of these four curves superimposed with the given points marked using the character * by application of plot. Same as Problem 14 but for the curves determined by Problems 9 through 12. Try Mathematica to obtain and compare results for the above problems.

LEASTSQG 1. Given four points (1,0.5), (2.5,0.88), (3.2,1.35), and (4.5,2.76), they are to be least-squares fitted by a linear combination of two selected functions, f1(x) = e-x and f2(x) = e–2x, in the form of f(x) = c1f1(x) + c2f2(x). Find c1 and c2. 2. For a given set of 10 points, (xi,yi) for i = 1,2,…,10, the least-squares method has been applied to fit these data by two students. Student A selects 3 functions f1(x) = x, f2(x) = x3, and f3(x) = x5 to obtain the coefficients a1, a2, and a3 for the expressiony = a1x + a2x3 + a3x5. Student B selects 3 other functions f1(x) = sinx, f2(x) = sin3x, and f3(x) = sin5x to obtain the coefficients b1, b2, and b3 for the expression y = b1sinx + b2sin3x + b3sin5x. Write a program to calculate the least-squares errors EA and EB for the curve-fit approaches taken by the students A and B, respectively. 3. Apply the least-squares method to curve-fit the following three given points by a linear combination of two selected functions f1(x) = x and f2(x) = x3, namely, y = c1x + c2x3: x y

0 –1

1 0

2 2

Find c1 and c2 by use of Cramer’s Rule. © 2001 by CRC Press LLC

4. Given eight data points (Xi,Yi) for i = 1,2,…,8 as listed, fit them by the least-squares method with the equation Y = a1 + a2X + a3X2. Find a1–3 by applying the computer program LeastSqG. X Y

1 1.13

2 1.45

3 1.76

4 2.19

5 2.43

6 2.79

7 3.51

8 4.88

5. A set of three points are provided as listed: (1,0.2), (2,0.5), and (3,0.6). These points are to be fitted by application of the least-squares method using a linear combination of (a) two functions x and x2, or (b) two other functions x0 and x3. Which fit will be better, a or b? Back up your answer with detailed calculations. 6. Given three points (1,2), (3,5), and (4,13), two selected functions f1(x) = x–1 and f2(x) = x3 are to be linearly combined to fit these points by the equation y = a1f1(x) + a2f2(x) = a1(x–1) + a2x3. Derive two equations needed for solving a1 and a2 by use of the Least-Squares method. 7. Given 7 points of which the coordinates are X(i) and Y(i) for i = 1 to 7, a least-squares fit of these points with a linear combination of 3 selected functions f1(X) = X, f2(X) = sin2X, and f3(X) = e-X in the form of Y(X) = C(1)f1(X) + C(2)f2(X) + C(3)f3(X) has been conducted and the coefficients C(1) to C(3) have been found. Complete the following segment of FORTRAN program to calculate the total error E which is the sum of the squares of the differences between Y(X(i)) and Y(i) for i = 1 to 7. DIMENSION X(7),Y(7),C(3) DATA X,Y,C/(17 real numbers separated by,)/ insert statements for < - - - calculation of E involving C, X, Y, and f1, f2, and f3. WRITE (*,*) E STOP END 8. For a given set of data (1,–2), (2,0), (3,1), and (4,3), two equations have been suggested to fit these points. They are Y = X–2 and Y = (-X2 + 7X–10)/2. Based on the least-squares criterion, which equation should be chosen to provide a better fit? Explain why. 9. Given 12 points of which the coordinates are X(i) and Y(i) for i = 1 to 7, a least-squares fit of these points with a linear combination of 4 selected functions f1(X) = X, f2(X) = sin2X, f3(X) = cos3X, and f4(X) = e-X in the form of Y(X) = C(1)f1(X) + C(2)f2(X) + C(3)f3(X) + C(4)f4(X) has been conducted and the coefficients C(1) to C(4) have been found. Complete the following segment of FORTRAN program to calculate the total error E which is the sum of the squares of the differences between Y(X(i)) and Y(i) for i = 1 to 12, using a FUNCTION subprogram F(I,X) which evaluate the Ith selected function at a specified X value for i = 1 to 4.

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10. Add to the graph obtained for Problem 9 by marking the data points with the character * and also a text ‘Cubic Spline of Temperature Data’ at an appropriate location not touching the spline curve.

2.7 REFERENCES 1. Y. C. Pao, “On Development of Engineering Animation Software,” in Computers in Engineering, K. Ishii, editor, ASME Publications, New York, 1994, pp. 851–855. 2. A. Ralston, A First Course in Numerical Analysis, McGraw-Hill, New York, 1965. 3. H. Flanders, R. R. Korfhage, and J. J. Price, A First Course in Calculus with Analytic Geometry, Academic Press, New York, 1973.

© 2001 by CRC Press LLC

3

Roots of Polynomials and Transcendental Equations

3.1 INTRODUCTION In the preceding chapter, we derive equations which fit a given of data either exactly, or, by using a criterion such as the least-squares method. Once such equations have been obtained in the form of y = C(x) when the data are two-dimensional, or, z = S(x,y) when the data are three-dimensional. It is next of common interest to find where the curve C(x) intercepts the x-axis, or, where the surface S(x,y) intercepts with the x-y plane. Mathematically, these are the problems of finding the roots of the equations C(x) = 0 and S(x,y) = 0, respectively. The equation to be solved could be a polynomial of the form P(x) = a1 + a2x + … + aixi–1 + … + aN + 1xN which is of Nth order, or, a transcendental equation such as C(x) = a1sinx + a2sin2x + a3sin3x. As it is well known, a polynomial of Nth order should have N roots which could be real, or, complex conjugate pair if the coefficients of the polynomial are all real. Geometrically speaking, only those real roots really pass the x-axis. For a transcendental equation, there may be infinite many roots. In this chapter, we shall introduce computational methods for finding the roots of polynomials and transcendental equations. Beginning with the very primitive approach of incremental and halfinterval searches, the approximate location of a particular root is to be located. More refined, systematic methods such as the linear interpolation and Newton-Raphson methods are then followed to determine the more precise location of the root. A program called FindRoot incorporating the four methods is to be presented for interactive solution of a particular root of a given polynomial or transcendental equation when the upper and lower bounds of the root are provided. Also discussed is a method called Successive Substitution. A transcendental equation derived from analysis of a four-bar linkage problem is used to demonstrate how roots are to be found by application of this method. Another transcendental equation has been derived for the unit-step response analysis of a mechanical vibration system and its roots solved by application of the Newton-Raphson method to illustrate how the design specifications are checked in the time domain. Since the Newton-Raphson method for solving F(x) = 0 which can be a polynomial, or, transcendental equation of one variable is based on the Taylor’s series involving the derivatives of F(x), it can be extended to the solution of two-equations F1(x,y) = 0 and F2(x,y) = 0 by application of Taylor’s series involving partial derivatives of both F1 and F2 with respect to x and y. A program called NewRaphG has been developed for this purpose. Also, this generalized Newton-Raphson method allows the quadratic factors of a higher order polynomial to be iteratively and continuously extracted and their quadratic roots solved by the so-called Bairstow method. For that, a program called Bairstow is made available for interactive application. © 2001 by CRC Press LLC

Both QuickBASIC and FORTRAN versions for the above-mentioned programs are presented. Both the application of the roots m-file of MATLAB in place of the program Bairstow and direct conversion of the program FindRoot into MATLAB version are also presented. The Mathematica’s function NSolve is introduced in place of the program Bairstow if the user prefers. Also the linear interpolation method used in the program FindRoot has been translated into Mathematica version. In fact. Mathematica has its own FindRoot based on the Newton-Raphson method.

3.2 ITERATIVE METHODS AND PROGRAM FINDROOT Program FindRoot is developed for interactive selection of an iterative method among the four made available: (1) Incremental Search, (2) Bisection Search, (3) Linear Interpolation, and (4) Newton-Raphson Iteration. Polynomials are often encountered in engineering analyses such as the characteristic equations in vibrational and buckling problems. The roots of a polynomial are related to some important physical properties of the systems being analyzed, such as the frequencies of vibration or buckling loads. A nth degree polynomial can be expressed as: P(x) = a1 + a 2 x + a 3x 2 + … + a n x n −1 + a n +1x n n +1

=

∑

a k x k −1 = 0

(1)

k =1

For n = 1,2,3, there are formulas readily available in standard mathematical handbooks1 for finding the roots. But for large n values, computer methods are then necessary to help find the roots of a given polynomial. The methods to be discussed here are simple and direct and are applicable to not only polynomials but also transcendental equations such as 5 + 7cosx – cos60° – cos(60° – x) = 0 related to a linkage design problem2 or x = 40000/{1–0.35sec[40(x/107)0.5]} arisen from buckling study of slender rods.3

INCREMENTAL SEARCH For convenience of discussion, let us consider a cubic equation: P(x) = 1 + 2 x + 3x 2 + 4 x3 = 0

(2)

To find a root of P(x), we first observe that P(x = –∞)<0, P(x = 0) = 1, and P(x = –∞)>0. This indicates that the P(x) curve must cross the x axis, possibly once or an odd number of times. Also, the curve may remain above the x-axis or cross it an even number of times. To further narrowing down the range on the x-axis, in which the root is located, we can begin to check the sign of P(x) at x = –10 and search toward the origin using an increment of x equal to 2. That is, we may construct a list such as:

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X P(x)

–10 –

–8 –

–6 –

–4 –

–2 –

0 +

Since P(x) changes sign from x = –2 to x = 0, this incremental search can be continued using an increment of x equal to 0.2 and the left bound x = –10 by replaced by x = –2 to obtain: X P(x)

–1.8 –

–1.6 –

• • • –

–0.8 –

–0.6 +

The search continues as follows: x P(x)

–0.78 –

–0.76 –

x P(x)

–0.618 –

x P(x)

–0.6058 +

–0.616 –

• • • – • • • –

–0.62 –

–0.60 +

–0.606 –

–0.604 +

If only three significant figures accuracy is required, then x = 0.606 is the root and it has taken 23 incremental search steps to arrive at this answer. If better accuracy is required, the root should then be sought between x = –0.6060 and x = –0.6058. Program FindRoot prepared both is QuickBASIC and FORTRAN has one of the options using the above-explained incremental search method, it also has other methods of finding the roots of polynomials and transcendental equations to be introduced next.

BISECTION SEARCH The above example of incremental search shows that if we search from left to right of the x-axis for the root of 4x3 + 3x2 + 2x + 1 = 0 between x = –2 and x = 0, it would be longer than if we search from right to left because the root is near x = 0. Rather than using a fixed incremental in the incremental search method, the bisectional method uses the mid-point of the two bounds of x in search of the root. It involves the testing of the signs of the polynomial at the bounds of the root and replacing the bounds. The two search methods follow the same procedure. So, the bisection method would go as follows: x P(x)

–10 –

x P(x)

–0.46875 +

© 2001 by CRC Press LLC

0 +

–5 –

–2.5 –

–0.546875 +

–1.25 –

–0.625 –

–0.5859375 +

–0.3125 +

–0.6054688 +

3. Find the roots of P(x) = x6 + x5–8x4 + 14x3 + 13x2–111x + 90 = 0 by application of the program Bairstow. Are the real roots graphically verified in the plot shown in Problem 2? 4. Apply root.m of MATLAB to find the roots of P(x) = x6 + x5–8x4 + 14x3 + 13x2 111x + 90 = 0. 5. Apply the program Bairstow to solve the characteristic equations derived by the program CharacEq: (1) 3–15 2–18 = 0, (2) 3–182 + 109 –222 = 0, and (3) 3–122 + 47 –60 = 0. 6. Apply the program Bairstow to solve for the characteristic equations derived for the program EigenODE.Stb: λ2 − 36λ + 243 = 0, λ3 − 96λ2 − 2560λ − 16384 = 0, λ4 − 200λ3 + 13125λ2 − 312500λ + 1953125 = 0, and λ5 − 360λ4 + 466563 − 2612736λ2 + 5.878656x10 7 λ − 3.62791x108 = 0 7. Apply roots.m of MATLAB to find the roots of the polynomials given in Problem 5 and then graphically verify the locations of these roots by plotting the polynomial curves using plot.m. 8. Repeat Problem 7 except for those polynomials given in Problem 6. 9. Apply NSolve of Mathematica to solve Problems 2 to 6.

3.6 REFERENCES 1. M. Abramowitz and I. A. Stegun, editors, “Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables,” Applied Mathematics Series 55, National Bureau of Standards, Washington, DC, 1964. 2. Y. C. Pao, Elements of Computer-Aided Design and Manufacturing, CAD/CAM, John Wiley & Sons, New York, 1984. 3. A. Higdon, E. H. Ohlsen, W. B. Stiles, J. A. Weese, and W. F. Riley, Mechanics of Materials, John Wiley & Sons, New York, Fourth Edition, 1985. 4. J. F. Traub, Iterative Methods for Solution of Equation, Prentice-Hall, Englewood Cliffs, NJ, 1964. 5. A. E. Taylor, Advanced Calculus, Ginn & Co., Boston, 1955. 6. W. H. Li, Fluid Mechanics in Water Resources Engineering, Allyn & Bacon, Boston, 1983. 7. C. R. Wylie, Jr., Advanced Engineering Mathematics, McGraw-Hill, New York, 1960.

© 2001 by CRC Press LLC

4

Finite Differences, Interpolation, and Numerical Differentiation

4.1 INTRODUCTION Linear interpolation is discussed in the preceding chapter as a method for finding a particular root of a polynomial, or, transcendental equation when the upper and lower bounds of the interval for search are provided. To continue the discussion of the general topic of interpolations which not necessarily linear could be quadratic (parabolic, cubic, quartic, and so on, we in this chapter present methods for this general need of interpolation in engineering analyses by treating not only equations but also a set of N tabulated data, (xi,yi) for i = 1–N. Finite difference table will be introduced and constructed for the equally-spaced data, that is x2–x1 = x3–x2 = … = xN-xN–1. This table can be utilized as a forward-difference, backward- difference, or, central-difference table depending on how its is applied for the interpolation use. Taylor’s series and a shifting operator are to be used in derivation of the interpolation formulas in terms of the forward-difference, backward-difference, and central-difference operators. A program DiffTabl has been developed for printing out a difference table of a set of equally-spaced data. Differentiation operator will also be introduced for the derivation of the numerical differentiation needs. When a set of equally-spaced data, (xi,yi) for i = 1–N, are given, formulas in terms of the forward-difference, backward-difference, and centraldifference operators are derived for the need of calculating the value of dy/dx at a listed x value or unlisted. If x is not equal to one of the xi, interpolation and differentiation have to be done combinedly through a modification of the Taylor’s series expansion. For curve-fit by polynomials and for interpolation, applications of the versatile Lagrangian interpolation formula are also discussed. A program called LagrangI is made available for this need. QuickBASIC, FORTRAN, and MATLAB versions of the above-mentioned programs are to be provided. Application of the Mathematica’s function Interpolating Polynomial in place of LagrangI is demonstrated. In solution of the problems governed by a system of ordinary differential equations with either some initial and/or boundary conditions specified, the finite differences will be applied. In Chapter 6, such method for finding the approximate answer to the problem is discussed. Accuracy of such approximate solution will depend on the increment of the independent variable, stepsize, adopted and on which approximate method is employed. © 2001 by CRC Press LLC

Because numerical differentiation is highly inaccurate, whenever possible numerical integration should be preferred over numerical differentiation. In case that one needs to find the velocity of a certain motion study and has the option of collecting the displacement or acceleration data, then the acceleration data should be taken not the displacement data. The reason is that one has the choice of applying numerical differentiation to the displacement data or numerical integration to the acceleration data to obtain the velocity results. The numerical integration which is the topic of Chapter 5 has the smoothing effect and hence is more accurate! Graphically, differentiation is of a local evaluation of determining the slope at a selected point on a curve which could be the result of fitting a number of data points discussed in Chapter 3 while integration is of a global evaluation of finding the area under the curve between two specified limits of the independent variable. For a set of three given points fitted linearly by two linear segments and quadratically by a parabola, the slope at the mid-point could have very different slope values while the areas under the linear segments and under the parabola would not differ too significantly. Hence, it is worthy of emphasizing that learning the computational methods is easier when compared to making decision of which method is best to solve the problem at hand.

4.2 PROGRAM DIFFTABL — APPLICATIONS OF FINITE-DIFFERENCE TABLE Program DiffTabl has been developed for the need of constructing a table of finite differences of a given set of N two-dimensional points, (xi,yi) for i = 1–N. The x values are assumed to be equally spaced, i.e., , x2–x1 = x3–x2 = ••• = xN-xN–1 = h, h being called the increment, or, stepsize. This so-called difference table can be applied for interpolation of the y value for a specified, unlisted x value inside the range of x = x1 and x = xN (extrapolation if outside the range), and differentiation. Table 1 shows a typical difference table. The symbol  used in Table 1 is called Forward Difference Operator. If we refer the numbers listed in the x and y columns as x1 to x6 and y1 to y6, respectively, the first number listed under y, 1.9495, is obtained from the calculation of y2–y1 and is identified as ∆y1. The last number listed in the y column, 5.3015, is equal to y6–y5 and referred to as y5. Or, we may write the general formula as, for i = 1 to 5, ∆y i = y i +1 − y i

(1)

yi is called the first forward difference of y at xi. The higher order forward differences listed in Table 1 are obtained by extended application of Equation 1. That is, ∆2 y i = ∆(y i +1 − y i ) = ∆y i +1 − ∆y i = y i +2 − 2 y i +1 + y i © 2001 by CRC Press LLC

(2)

TABLE 1 Difference Table (y = 1 to 2x + 3x2 to 4x3 + 5x4). x

y

1.1

4.4265

1.2

6.3760

y

 2y

 3y

 4y

 5y

1.9495 0.637 2.5865 1.3

8.9625

0.126 0.763

3.3495 1.4

12.3120

1.5

16.5625

1.6

21.8640

0.012 0.138

0.901 4.2505

0.000 0.012

0.150 1.051

5.3015

∆3y i = ∆2 (y i +1 − y i ) = ∆2 y i +1 − ∆2 y i = (y i +3 − 2 y i +2 + y i +1 ) − (y i +2 − 2 y i +1 + y i )

(3)

= y i +3 − 3y i +2 + 3y i +1 − y i and so on. We shall show later how the third through seven columns of Table 1 can be interpreted differently when the backward and central difference operators are introduced. First, we will demonstrate how Table 1 can be applied for interpolation of the y value at an unlisted x value, say y(x = 1.24). To do that, the shifting operator, E, needs to be introduced. The definition of E is such that: Ey i = y i +1

(4)

That is, if E is operating on yi, the y value is shifted down to the next provided y value. Interpolation is a problem of not shifting a full step but a fractional step. For the need of finding y at x = 1.24, the x value falls between x2 = 1.2 and x3 = 1.3. Since the stepsize, h, is equal to 0.1, a full shift from y2 = 6.3760 would lead to y3 which is equal to 8.9625. We expect the value of y(x = 1.24) to be between y2 and y3. Instead of E1y2, the value of E0.4y2 is to be calculated by shifting only 40%. To find the meaning of E0.24, or, more generally Er for 0
© 2001 by CRC Press LLC

or,

E = 1+ ∆

(5,6)

By application of binomial expansion, we can then have: E r = (1 + ∆ ) = r

∞

∑ ( )∆ r k

k

(7)

k =0

where the binomial coefficients are defined as:

( ) =1 r 0

( )= r k

and

r( r − 1)…[ r − ( k − 1)] 1⋅ 2L k

(8)

We can now use Equation 7 to obtain: y(x = 1.24) = E 0.4 y(x = 1.2) = E 0.4 y 2 = (1 + ∆ ) y 2 0.4

0.4(0.4 − 1) 2   = 1 + 0.4 ∆ + ∆ + … y 2 1⋅ 2  

(

(9)

)

= 1 + 0.4 ∆ − 1.12 ∆2 + 0.064 ∆3 − 0.0416 ∆4 + 0.022952 ∆5 y 2 Equation 9 can be applied for linear interpolation if up to the y2 terms are adopted; for parabolic interpolation if up to the 2y2 terms are adopted; and so on. Since Table 1 has up to the fifth order forward differences available but the last column contains a zero value, Equation 9 can therefore be effectively up to the fourth-order forward-difference interpolation. The numerical results of y(x = 1.24) using linear, parabolic, cubic, and fourth-order are 7.4106, 7.3190, 7.3279, and 7.3274, respectively. Since we know y = 1–2x + 3x2–4x3 + 5x4, the exact value of y(x = 1.24) is equal to 7.3274. An explanation for discrepancies in all of these four attempts of interpolations, relative to the exact value, is provided in a homework exercise given in the Problems set.

BACKWARD-DIFFERENCE OPERATOR Notice that the first numbers listed in columns three through seven in Table 1 are the five forward differences of y1, and that only four forward differences (the second numbers in columns three through six) of y2 are available. Lesser and lesser forward differences are available for later y’s until there is only y5 for y5. That is to say, to interpolate y(x) for an x value between x5 = 1.5 and x6 = 1.6, Equation 9 can only be used up to the y5 term. To remedy this situation and to make most use of the provided set of 6 (x,y) data, it is appropriate at this time to introduce the backward-difference operator, , which is defined as: ∇y i = y i − y i −1 © 2001 by CRC Press LLC

(10)

By combining Equations 1, 7, and 10, we notice that: ∇y i +1 = y i +1 − y i = ∆y i

(11)

and

(

)

∇y i −1 = y i +1 − y i = 1 − E −1 y i +1

(12)

So, ∇ = 1 − E,−1

or,

E −1 = 1 − ∇

(13,14)

Equation 12 is an important result because it indicates that the last numbers listed in columns three through seven of Table 1 are the first through five backward differences of y6. If we could derive an interpolation formula in terms of , there are up to fifth-order backward difference of y6 available. Toward that end, let us consider the need of interpolating the value of y(x = 1.56). This y value can be reached by shifting backward by 0.4 step from x = 1.6 since the stepsize for Table 1 is h = 0.1. By using Equation 14 and noticing Equations 7 and 8, we can have: y(x = 1.56) = E −0.4 y(x = 1.6) = E −0.4 y 6 = (1 − ∇) y 6 0.4

0.4(0.4 − 1)  = 1 + 0.4( −∇) + (−∇)2 + …y6 1⋅ 2  

(

)

= 1 − 0.4∇ − 0.12∇2 − 0.064∇3 − 0.0416∇ 4 − 0.02295∇5 y 6 (15) One can then apply Equation 15 to obtain the interpolated y(x = 1.56) values using up to the fifth order backward differences. This is left as a homework exercise given in the Problems set.

CENTRAL-DIFFERENCE OPERATOR For the interest of completeness and later application in numerical solution of ordinary differential equations, we also introduce the central difference operator, . When it is operating on yi, the definition is: h δy i = y x i +  − y x i −   2

h 2

(16)

The first-order central difference is seldom used and the second-order central difference is frequently applied, which is: h   h   δ 2 y i = δ y x i +   − δ y x i −   = y(x i + h ) − 2 y(x i ) + y(x i − h )    2   2  = y i +1 − 2 y i + y i −1 © 2001 by CRC Press LLC

(17)

DIFFERENTIATION OPERATOR Another important operator that needs to be introduced in connection with the application of difference table is the differentiation operator, D, which is defined as: Dy(x = x i ) ≡

dy(x) dx

x= xi

≡ Dy i

(18)

As it is our intention to apply an available difference table for numerical differentiation at one of the listed x values, or, at an unlisted x value, by using either the forward or backward differences of y values. For example, we may want to find Dy at x = 1.2, or, at x = 1.24. To derive an expression for D in terms of , we recall the Taylor’s series of a function y(x = a + h) near the neighborhood of x = a for a small increment of h: y(a + h ) = y(a ) +

y ′( a ) y′′(a ) 2 y ( j ) (a ) j h+ h +…+ h +… j! 1! 2!

(19)

where y(j) is the jth derivative with respect to x. Using the notation of differentiation operator D and the shifting operator E, the above expression can be written as: Ey(a ) = y(a ) +

hDy(a ) h 2 D2 y(a ) h jD jy(a ) + +…+ +… j! 1! 2!

[

(20)

]

= 1 + hD + h 2 D2 + … + h jD j + … y(a ) = e hD y(a ) or, E = e hD

and

D=

1 lnE h

(21)

In order to use the difference table for numerical differentiation, we substitute Equation 6 into Equation 21 to obtain: D=

1 ln(1 + ∆ ) h

(22)

By substituting the logarithmic function in Equation 21 with an infinite series1 and applying the D operator for yi, the result is: Dy i =

1 1 1 ∆ − ∆2 + ∆3 − … y i  h 2 3

(23)

Hence, to find Dy(x = 1.2) by using the finite differences in Table 1, Equation 23 can be applied to obtain: © 2001 by CRC Press LLC

Dy 2 =

1  1 1 2.5865 − 0.5x 0.763 + 0.138 − 0.012 = 22.48  3 4 0.1 

Notice that the above result is when up to the fourth-order forward differences of y2 are all utilized. Linear, parabolic, and cubic numerical differentiations at x = 1.2 could also be calculated by taking only one, two, and three terms inside the parentheses of the above expression. The respective results are 25.865, 22.05, and 22.51. Since y(x) = 1–2x + 3x2–4x3 + 5x4 and y(x) = –2 + 6x–12x2 + 20x3, the exact value of y(x = 1.2) = –2 + 7.2–17.28 + 34.56 = 22.48 indicates that the fourth-order calculation is the best. When Dyi is needed for xi near the end of x list, it is better to express D in terms of the backward-difference operation , which based on Equations 14 and 21 is: Dy(x i ) =

[

]

1 1 (ln E)y i = ln(1 − ∇)−1 y i h h

 −1  ∇ 2 ∇3 ∇ 4 1 1 1 = −∇ − ∇2 − ∇3 − … y i =  ∇ + + + + … y i  h  h 2 3 2 3 4 

(24)

The shifting operator E and differentiation operator D can be combined to derive formulas for numerical differentiation of y(x) at x values unlisted in the difference table either in terms of forward-difference operator or backward-difference. First, let recall Equations 7 and 23 and apply them to find y(x = xi + rh) in terms of the forward-difference operator as follows: y′(x i + rh ) =

1 1 4 DE r y i = ln(1 + ∆ )(1 + ∆ ) y i h h

=

 ∆2 ∆3 r( r − 1) 2 1  ∆ + … y i + − … 1 + r∆ + ∆ − h 2 3 2  

=

 1 2 r − 1 2 3r 2 − 6 r + 2 3 2 r 3 − 9 r 2 + 11r − 3 4 ∆ + ∆ + ∆ + … y i ∆ + h 2 6 12 

(25)

Similarly, y(xi-rh) can be expressed in terms of backward-differential operator, as: y′(x i − rh ) =

1 1 r DE − r y i = [− ln(1 − ∇)](1 − ∇) y i h h

=

 r( r − 1) 2 ∇ 2 ∇3 1  + + … 1 − r∇ + ∇ + … y i ∇ + h 2 3 2  

=

 1 2 r − 1 2 3r 2 − 6 r + 2 3 2 r 3 − 9 r 2 + 11r − 3 4 ∇ + … y i ∇ + ∇ − ∇ − h 12 2 6 

© 2001 by CRC Press LLC

(26)

Input[1]: = Y = {2,4,7,11,24}; Input[2]: = <
x x x x x x

x x x x x x

x x x x x x

x x x x x x

x x x x x x

Input[3]: = Do[yanddys[[i,1]] = y[[i]]; MatrixForm[yanddys] Output[3] = 1 3 7 12 44 78

x x x x x x

x x x x x x

x x x x x x

x x x x x x

x x x x x x

Input[4]: = Do[Do[yanddys[[i,j]] = yanddys[[i + 1,j–1]]-yanddys[[i,j–1], {i,n-j + 1}],{j,2,n}]; MatrixForm[yanddys] Output[4] = 1 3 7 12 44 78

2 4 5 32 34 x

© 2001 by CRC Press LLC

2 –1 27 –78 1 26 –51 x 27 –25 x x 2 x x x x x x x x x x x

Notice that plot.m automatically uses solid, broken, and dotted lines to plot the four-order polynomial curve based on arrays XC and YC, and the vertical line based on arrays XV, YV, and the horizontal line based on arrays XH, YH, respectively. The details involved in exact curve-fit of the five given point by applying LeastSqG.m already has been discussed in the program Gauss. The coefficients, {C}, of the fourth-order polynomial determined by LeastSqG.m are arranged in descending order. In order to apply polyval.m of MATLAB, the order of {C} has to be reversed and stored in {Creverse} which is implemented above by the for and end loop.

MATHEMATICA APPLICATIONS Derivation of the polynomial which passing through a set of given (x,y) points based on the Lagrangian formula can be achieved by application of the Interpolating Polynomial function of Mathematica. For example, a fourth-order polynomial can be derived for a given set of 5 (x,y) data points as follows: In[1]: = pofx = InterpolatingPolynomial [{{1,2},{2,4},{3,6},{4,8},{5,11}},x]

(−4 + x) (−3 + x) (−2 + x)  −1 + x  Out[1]: = 2 +  2 + ) (   24 To interpolate the y value of using the derived polynomial at x equal to 4.56, we replace all x’s appearing in the above expression (saved in pofx) with a value of 4.56 by interactively entering In[2]: = pofx/. x -> 4.56 Out[2]: = 9.45174 Linear and parabolic interpolations can also be implemented by selecting appropriate data points from the given set. For example, to interpolate the y value at x = 1.25 by linear interpolation, we enter: In[3]: = p1 = InterpolatingPolynomial[{{1,2},{2,4}},x] Out[3]: = 2 + 2 (–1 + x) In[4]: = p1/. x -> 1.25 Out[4]: = 2.5 To parabolically interpolate the y value at x = 3.75 using the points (3,6), (4,8), and (5,11), the interactive application of Mathematica goes as: In[5]: = p2 = InterpolatingPolynomial[{{3,6},{4,8},{5,11}},x]

© 2001 by CRC Press LLC

Out[5]: = 6 +  2 + 

−4 + x  (−3 + x) 2 

In[6]: = p2/. x -> 3.75 Out[6]: = 7.40625

4.4 PROBLEMS DIFFTABL 1. Construct the difference table based on the following listed data and then find the y value at x = 4.5 by using the backward-difference formula up to the third-order difference. x y

1 2

2 4

3 7

4 12

5 20

2. Explain why interpolations using Equation 9 by the first through fourth orders all fail to match the exact value of y(x = 1.24) = 7.3274 by making 4 plots for x values ranging from 1.2 to 1.3 with an increment of x = 0.001. These 4 plots are to be generated with the 4 equations obtained when the first 2, 3, 4, and 5 points are fitted by a first-, second-, third-, and fourth-degree polynomials, respectively. Also, draw a x = 1.24, vertical line crossing all 4 curves. 3. Find the first-, second-, third-, and fourth-order results of y(x = 1.56) by use of Equation 15. 4. Write Er in terms of binomial coefficient and the backward-difference operator , similar to Equation 7. 5. Find the first-, second-, third-, and fourth-order results of y(x = 1.24) by use of Equation 24. 6. Find the first-, second-, third-, and fourth-order results of y(x = 1.56) by use of Equation 25. 7. Given 6 (x,y) points (1,0.2), (2,0.4), (3,0.7), (4,1.5), (5,2.9), and (6,4.7), parabolically interpolate y(x = 3.4) first by use of forward differences and then by use of backward differences. 8. Modify either the QuickBASIC or FORTRAN version of the program DiffTabl to include the fifth difference for the need of forward or backward interpolation and numerical differentiation. 9. Given 5 (x,y) points (0,0), (1,1), (2,8), (3,27), and (4,64), construct a complete difference table based on these data. Compute (1) y value at x = 1.25 using a forward, parabolic (second-order) interpolation, (2) y value at x = 3.7 using a backward, cubic (third-order) interpolation, and (3) dy/dx value at x = 0 using a forward, third-order approximation. 10. Based on Equation 21, derive the forward-difference formulas for D2yi and D3yi. 11. Use the result of Problem 10 to compute D2y2 and D3y1 by adopting the forward-difference terms in Table 1 as high as available. © 2001 by CRC Press LLC

12. Use the data in Table 1 to compute the first derivative of y at x = 1.155 by including terms up to the third-order forward difference. 13. Apply MATLAB for the points given in Problem 1 to print out the rows of x, y, ∆y, 2y, 3y, and 4y. 14. Same as Problem 13 but the points in Problem 6. 15. Apply Mathematica and DO loops to print out a difference table similar to that shown in Mathematica Application of Section 4.2 for the points given in Problem 1. 16. Apply Mathematica and DO loops to print out a difference table similar to that shown in Mathematica Application of Section 4.2 for the points given in Problem 6. 17. Compute the binomial coefficient for r = 0.4 and k = 1,2,3,4,5 according to Equation (8) in Section 4.2 using MATLAB. 18. Rework Problem 17 but using Mathematica.

LAGRANGI 1. Given five points (1,1), (2,3), (3,2), (4,5), and (5,4), use the last three points and Lagrangian interpolation formula to compute y value at x = 6. 2. A set of 5 (x,y) points is given as (1,2), (2,4), (4,5), (5,2), (6,0), apply the Lagrangian interpolation formulas to find the y for x = 3 by parabolic interpolation using the middle three points. Check the answer by (a) without fitting the three points by a parabolic equation, and (b) by deriving the parabolic equation and then substituting x equal to 3 to find the y value. 3. Apply the Lagrangian formula to curve-fit the following listed data near x = 5 by a cubic equation. Use the derived cubic equation to find the y value at x = 4.5. x y

1 2

2 4

3 7

4 12

5 20

4. Use the data set given in Problem 3 to exactly curve-fit them by a quartic equation y(x) = a1 + a2x + a3x2 + a4x3 + a5x4. Do this manually based on the Lagrangian formula. 5. Write a program and call it ExactFit.Ln5 for computation of the coefficients a1–5 in the y(x) expression in Problem 4. 6. Generalize the need in Problem 4 by extending the exact fit of N given (x,y) points by a polynomial y(x) = a1 + a2x + … + aixi–1 + … + aNxN–1 based on the Lagrangian formula. Call this program ExactFit.LnN. 7. Based on the Lagrangian formula, use the first four of the five points given in Problem 1 to interpolate the y value at x = 2.5 and then the last four of the five points also at x = 2.5. 8. Write a program and call it Expand.1 which will expand the set of five points given in Problem 2 to a set of 21 points by using an increment of x equal to 0.2 and linear interpolation based on the Lagrangian formula. For any x value which is not equal to any of the x values of the five given © 2001 by CRC Press LLC

9. 10.

11. 12. 13. 14. 15.

points, this x value is to be tested to determine between which two points it is located. These two given points are to be used in the interpolation process by setting N equal to 2 in the program LagrangI. This procedure is to be repeated for x values between 1 and 6 in computation of all new y values. As for Problem 8 except parabolic interpolation is to be implemented. Call the new program Expand.2. Extend the concept discussed in Problems 8 and 9 to develop a general program Expand.M for using N given points and Mth-order Lagrangian interpolation to obtain an expanded set. Apply the function InterpolatingPolynomial of Mathematica to solve Problems 1 and 2. Check the result of Problem 4 by Mathematica. Apply LagrangI.m to solve Problem 1 by MATLAB. Apply LagrangI.m to solve Problem 2 by MATLAB. Apply LagrangI.m to solve Problem 7 by MATLAB.

4.5 REFERENCE 1. R. C. Weast, Editor-in-Chief, CRC Standard Mathematical Tables, the Chemical Rubber Co. (now CRC Press LLC), Cleveland, OH, 1964, p. 381.

© 2001 by CRC Press LLC

MATHEMATICA APPLICATION Mathematica numerically integrate a function f(x) over the interval x = a to x = b by use of the function NIntegrate. The following example demonstrates the computation of one quarter of a circle having a radius equal to 2: In[1]: = NIntegrate[Sqrt[4x^2], {x, 0, 2}] Out[1] = 3.14159

5.3 PROGRAM VOLUME — NUMERICAL APPROXIMATION OF DOUBLE INTEGRATION Program Volume is designed for numerical calculation of double integration involving an integrand function of two variables. For convenience of graphical interpretation, the two variables x and y are usually chosen and the integrand function is denoted as z(x,y). If the double integration is to be carried for the region xL≤x≤xU and yL≤y≤yU, the value to be calculated is the volume bounded by the z surface, z = 0 plane, and the four bounding planes x = xL, x = xU, y = yL, and y = yU where the sub-scripts L and U are used to indicate the lower and upper bounds, respectively. The rectangular region xL≤x≤xU and yL≤y≤yU on the z = 0 plane is called the base area. The volume is there-fore a column which rises above the base area and bounded by the z(x,y) surface, assuming that z is always positive. Mathematically, the volume can be expressed as:

V=∫

yU

yL

∫

xU

xL

z ( x , y )d

(1)

If we are interested in finding the volume of sphere of radius equal to R, the bounds can be selected as xL = yL = 0 and xU = yU = R, and let z = (R2x2y2).5. Equation 1 can then be applied to find the one-eighth of spherical volume. In fact, the result can be obtained analytically for this z(x,y) function. We are here, however, interested in a computational method for the case when the integrand function z(x,y) is too complex to allow analytical solution. The trapezoidal rule for single integration discussed in the program NuIntgra can be extended to double integration by observing from Figure 1 that the total volume V can be estimated as a sum of all columns erected within the space bounded by the z surface and the base area. In Figure 5, the integrand functions used are:

(

z(x, y) = R 2 − x 2 − y 2

)

0.5

, for x 2 + y 2 ≤ R 2

(2)

for x 2 + y 2 > R 2

(3)

and z(x, y) = 0,

© 2001 by CRC Press LLC

Vi, j =

(

)

1 z +z +z +z ∆x∆y 4 i, j i, j+1 i +1, j i +1, j+1

(6)

where:

(

z i, j ≡ z x i , y j

)

(7)

The total volume is then the sum of all Vi,j for i ranging from 1 to M and j ranging from 1 to N. Or, V=

R

∫∫ 0

=

R

0

∆x∆y 4

(x

2

)

.5

+ y 2 dxdy

∑ ∑( i

z i, j + z i, j+1 + z i +1, j + z i +1, j+1

j

)

(8)

The two summations in Equation 8 are loosely stated. Actually, the heights calculated at all MxN grid-points on the base area used in Equation 7 can be separated into three groups: (1) those heights at the corners whose coordinates are (0,0), (0,R), (R,0), and (R,R), are needed only once, (2) those heights on the edges of the base area, excluding those at the corners, are needed twice because they are shared by two adjacent columns, and (3) all heights at interior grid-points are needed four times in Equation 8 because they are shared by four adjacent columns. That is to say, in terms of the subscripts I and j the weighting coefficients, wi,j, for zi,j can be summarized as follows: wi,j = 1 = 4 = 2

for for for

(i,j) = (1,1),(1,N),(M,1),(M,N), i = 2,3,…,M-1 and j = 2,3,…,N-1 other i and j combinations

Subsequently, Equation 8 can be written as: V=

∆x∆y 4

M

N

i =1

j=1

∑∑w

z

i, j i, j

(9)

A more precise way to express V in terms zi,j is to introduce a weighting coefficient vector for Trapezoidal rule, {wt}. Since we have averaged the four heights of each contributing column, that is linearly connecting the four heights. That is, the trapezoidal rule has been applied twice, one in x-direction and another in ydirection. When M and N stations are employed in x- and y-directions, respectively. we may therefore define two weighting coefficient vectors

© 2001 by CRC Press LLC

{w } = [1

2 2 … 2 2 2]

T

(10)

{w } = [1

2 2 … 2 2 1]

T

(11)

t x

and t y

It should be noted that the subscripts x and y are added to indicate their association with the x- and y-axes, respectively, and that the orders of these two vectors are M and N, respectively, and that the beginning and ending components in both vectors are equal to one and the other components are equal to two. Having defined {wt}x and {wt}y, it is now easy to show that Equation 9 can be simply written as: V=

T ∆x {wt }x [Z]{wt }y ∆2y 2

(12)

where [Z] is a matrix of order M by N having zi,j as its elements. Since {wt}x is a vector of order M by one, its transpose is of order one by M and {wt}y is of order N by one, the matrix multiplication of the three matrices can be carried out and the result does agree with the requirement on wi,j spelled out in Equation 9. Use of weighting coefficient vectors has the advantage of extending the numerical evaluation of double integrals from trapezoidal rule to Simpson’s rule where three adjacent heights are parabolically fitted (referring to program NuIntgra for more details). To illustrate this point, let us first introduce a weighting coefficient vector for Simpson’s rule as:

{w } = [1 s

4 2 … repeat of 4 and 2 … 4 1]

T

(13)

If we wish to integrate by application of Simpson’s rule in both x- and ydirections and using M and N (both must be odd) stations, the formula for the volume is simply: V=

T ∆x {ws}x [Z]{ws}y ∆3y 3

(14)

If for some reason one wants to integrate using Simpson’s rule along x-direction by adopting M (odd) stations, and using trapezoidal rule along y-direction by adopting N (no restriction whether odd or even) stations, then: V=

T ∆x {ws}x [Z]{wt }y ∆2y 3

(15)

Let us present a numerical example to further clarify the above concept of numerical volume integration. Consider the problem of estimating the volume

© 2001 by CRC Press LLC

Input]: = Sphere = Plot3D[If[4X^2Y^2>0, Sqrt[4X^2Y^2],0, {X,–2,2},{Y,–2,2},PlotPoints->{60,60}] The If command tests the first expression inside the brackets, it the condition is true then the statement which follows is implemented and other the last statement inside the bracket is implemented. In this case, the surface only rises over the base circle of radius equal to 2. The PlotPoints command specifies how many gird points along X- and Y-directions should be taken to plot the surface. The default number of point is 15 in both directions. The greater the number of grid points, the smoother the surface looks. The same result can be obtained by first defining a surface function, say sf, and then apply Plot3d for drawing the surface using sf as follows: Input]: = sf[X_,Y_] = If[4X^2Y^2>0, Sqrt[4X^2Y^2], 0] Input[2]: = Plot3D[sf[X,Y],{X,–2,2},{Y,–2,2},PlotPoints->{60,60}]

5.4 PROBLEMS NUINTGRA 1. Having learned how to apply Trapezoidal Rule for numerical integration, how would you find the area under the line y(x) = 1 + 2x and between x = 1 and x = 2? Do it not by direct integration, but numerically. What should be the stepsize for x in order to ensure an accurate result? 2. Having learned how to apply Simpson’s Rule for numerical integration, how would you find the area under the parabolic curve y(x) = 1 + 2x + 3x2 and between x = 1 and x = 2? Do it not by direct integration but numerically! What should be the stepsize for x in order to ensure an accurate result? 3. If Trapezoidal Rule, instead of Simpson’s Rule, is applied for Problem 2, find out how small should be the stepsize for x in order to achieve the same result accurate to the fifth significant digit. 4. Could Simpson’s Rule be applied for Problem 1? Would the result be different? If the result is the same, explain why. 5. Given five points (1,1), (2,3), (3,2), (4,5), and (5,4), use a stepsize of x = 1 to compute ydx by application of Simpson’s and Trapezoidal rules. 6. Use the trapezoidal and Simpson’s rules to find the area within the ellipse described by the equation (x/a)2 + (y/b)2 = 1. Compare the numerical results with the exact solution of ab. 7. Implement the integration of the function f(x) = 3e–2xsinx over the interval from x = 0 to x = 1 (in radian) by applying both the Trapezoidal and Simpson’s rules and using an increment of x = 0.25. 8. Find the exact solution of Problem 7 by referring to an integration formula for f(x) from any calculus book. Decrease the increment of x (i.e., , © 2001 by CRC Press LLC

9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

increase the number of points at which the integrand function is computed) to try to achieve this analytical result using both Trapezoidal and Simpson’s Rules. Apply the function Quad.m of MATLAB to solve Problem 1. Apply the function Quad.m of MATLAB to solve Problem 2. Apply the function Quad.m of MATLAB to solve Problem 6. Apply the function Quad.m of MATLAB to solve Problem 7. Apply MATLAB to spline curve-fit the five points given in Problem 5 and then integrate. Apply the function NIntegrate of Mathematica to solve Problem 1. Apply the function NIntegrate of Mathematica to solve Problem 2. Apply the function NIntegrate of Mathematica to solve Problem 6. Apply the function NIntegrate of Mathematica to solve Problem 7. Problem 13 but apply Mathematica instead.

VOLUME 1. Apply trapezoidal rule for integration along the x direction and Simpson’s rule along the y direction to calculate the volume under the surface z(x,y) = 3x + 2y2 + 1 over the rectangular region 0≤x≤2 and 0≤y≤4 using increments x = y = 1. 2. Rework Problem 1 except trapezoidal rule is applied for both x and y directions. 3. Find by numerical integration of the ellipsoidal volume based on the double integral 3[1–(x/5)2–(y/4)2]1/2dxdy and for x values ranging from 2 to 4 and y values ranging from 1 to 2. Three stations (for using Simpson’s rule) for the x integration and two stations (for using trapezoidal rule) for the y integration are to be adopted. 4. Find the volume between the z = 0 plane and the spherical surface z(x,y) = [4x2 – y2]1/2 for x and y both ranging from 0 to 2 by applying the Simpson’s rule for both x and y integrations. Three stations are to be taken along the x direction and five stations along the y direction for the specified numerical integration. 5. Specify a FUNCTION Z(x,y) for program Volume so that the volume enclosed by the ellipsoid (x/a)2 + (y/b)2 + (z/c)2 = 1 can be estimated by numerical integration and compare to the exact solution of 4abc/3.2 6. In Figure 8, the shape and dimensions of a pyramid are described by the coordinates of the five points (Xi,Yi) for I = 1,2,…,5. For application of numerical integration to determine its volume by either trapezoidal or Simpson’s rule, we have to partition the projected plane P2P3P4P5 into a gridwork. At each interception point of the gridwork, (X,Y), the height Z(X,Y) needs to be calculated which requires knowing the equations describing the planes P1P2P3, P1P3P4, P1P4P5and P1P5P2. The equation of a plane can be written in the form of 2(X–a) + m(Y–b) + n(Z–c) = 1 where (a,b,c) is a point on the plane and (2,m,n) are the directional cosines © 2001 by CRC Press LLC

6

Ordinary Differential Equations — Initial and Boundary Value Problems

6.1 INTRODUCTION An example of historical interest in solving an unknown function which is governed by an ordinary differential equation and an initial condition is the case of finding y(x) from: dy =y dx

y(x = 0) = 1

and

(1)

As we all know, y(x) = ex. In fact, the exponential function ex is defined by an infinite series: ex = 1 +

x1 x 2 + +… = 1! 2!

∝

∑ i=0

xi i!

(2)

To prove that Equation 2 indeed is the solution for y satisfying Equation 1, here we apply an iterative procedure of successive integration using a counter k. First, we integrate both sides of Equation 1 with respect to x:

∫

x

0

dy dx = dx

x

∫ y dx

(3)

0

After substituting the initial condition of y(x = 0) = 1, we obtain:

∫

x

y(x) = 1 + y dx

(4)

0

So we are expected to find an unknown y(x) which is to be obtained by integrating it? Numerically, we can do it by assuming a y(x) initially (k = 1) equal to 1, investigate how Equation 4 would help us to obtain the next (k = 2), guessed y(x), and hope eventually the iterative process would lead us to a solution. The iterative equation, therefore, is for k = 1,2,…

© 2001 by CRC Press LLC

∫

x

y( k +1) (x) = 1 + y( k )dx

(5)

0

The results are y(1) = 1, y(2) = 1 + x, y(3) = 1 + x + (x2/2!), and eventually the final answer is the infinite series given by Equation 2. What really need to be discussed in this chapter is not to obtain an analytical expression for y(x) by solving Equation 1 and rather to compute the numerical values of y(x) when x is equally incremented. That is, for a selected x increment, x (or, stepsize h), to find yiy(xi) for i = 1,2,… until x reaches a terminating value of xe and xi = (i–1) x. A simplest method to find y2 is to approximate the derivative of y(x) by using the forward difference at x1. That is, according to the notation used in Chapter 4, we can have: dy dx

x = x1

=˙

∆y1 y 2 − y1 = = y1 ∆x ∆x

(6)

Or, y2 = (1 + x)y1. In fact, this result can be extended to any xi to obtain yi + 1 = 1 + (x)yi. For the general case when the right-hand side of Equation 1 is equal to a prescribed function f(x), we arrive at the Euler’s formula yi + 1 = yi + f(xi) x. Euler’s formula is easy to apply but is inaccurate. For example, using a x = 0.1 in solving Equation 1 with the Euler’s formula, it leads to y2 = (1 + 0.1)y1 = 1.1, y3 = (1 + 0.1)y2 = 1.21 when the exact values are y2 = e0.1 = 1.1052 and y3 = e0.2 = 1.2214, respectively. The computational errors accumulate very rapidly. In this chapter, we shall introduce the most commonly adopted method of RungeKutta for solution of the initial-value problems governed by ordinary differential equation(s). For the fourth-order Runge-Kutta method, the error per each computational step is of order h5 where h is the stepsize. Converting the higher-order ordinary differential equation(s) into the standardized form using the state variables will be illustrated and computer programs will be developed for numerical solution of the problem. Engineering problems which are governed by ordinary differential equations and also some associated conditions at certain boundaries will be also be discussed. Numerical methods of solution based on the Runge-Kutta procedure and the finitedifference approximation will both be explained.

6.2 PROGRAM RUNGEKUT — APPLICATION OF THE RUNGE-KUTTA METHOD FOR SOLVING THE INITIAL-VALUE PROBLEMS Program RungeKut is designed for solving the initial-value problems governed by ordinary differential equations using the fourth-order Runge-Kutta method. There are numerous physical problems which are mathematically governed by a set of ordinary differential equations (ODE) involving many unknown functions. These unknown functions are all dependent of a variable t. Supplementing to this set of © 2001 by CRC Press LLC

Out[1] = {{X1 -> InterpolatingFunction[{0., 25.}, <>], X2 -> InterpolatingFunction[{0., 25.}, <>]}} In[1] shows that NDSolve has three arguments: the first argument defines the two ordinary differential equations and the initial conditions, the second argument lists the dependent variables, and the third argument specifies the independent variable and the range of investigation. Id is a name selected for easy later reference of the results generated. Out[1] indicates that interpolation functions have been generated for X1 and X2 for t in the range from 0 to 25. To print the values of X1 and X2, Id can be referred to interpolate the needs as follows: In[2]: = (Do[Print["t =", tv," X1 =", X1[tv]/. Id, " X2 =", X2[tv]/. Id], {tv, 0, 25, 1}]) Out[2] = t = 0 t= t= t= t= t= t= t= t= t= t= t= t= t= t= t= t= t= t= t= t= t= t= t= t= t=

© 2001 by CRC Press LLC

X1 = {0.} 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 =

X2 = {0.} {0.13827} {0.267432} {0.280915} {0.256722} {0.245409} {0.246925} {0.249969} {0.250675} {0.250238} {0.249931} {0.249923} {0.24999} {0.250014} {0.250006} {0.249999} {0.249998} {0.25} {0.25} {0.25} {0.25} {0.25} {0.25} {0.25} {0.25} {0.25}

X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 =

{0.183528} {0.0629972} {–0.0193286} {–0.0206924} {–0.00278874} {0.00365961} {0.00187841} {–0.000172054} {–0.000477574} {–0.000125015} {0.0000639464} {0.0000489215} {4.83211 10–7} {–0.0000104109} {–3.5685 10–6} {1.38473 10–6} {1.35708 10–6} {1.39173 10–7} {–4.40766 10–7} {–4.61875 10–7} {–1.93084 10–8} {2.47763 10–8} {9.81364 10–8} {6.7364 10–8} {3.57288 10–8}

These results are in agreement with the plotted curves shown in the MATLAB application. In[2] shows that the replacement operator/. is employed in X1[tv]/.Id which requires all t appearing in the resulting interpolating function for X1(t) created in the Id statement to be substituted by the value of tv. The looping DO statement instructs that the tv values be changed from a minimum of 0 and a maximum of 25 using an increment of 1. To have a closer look of the overshoot region, In[2] can be modified to yield In[3]: = (Do[Print[“t = “,tv,” X1 = “,X1[tv]/. Id,” X2 = “,X2[tv]/. Id], {tv, 2, 3, 0.1}]) Out[3] =

t=2 t= t= t= t= t= t= t= t= t= t=

X1 = {0.267432} 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.

X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 = X1 =

{0.273097} {0.277544} {0.280862} {0.283145} {0.284496} {0.285019} {0.284819} {0.284002} {0.282669} {0.280915}

X2 = {0.0629972} X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 = X2 =

{0.0504262} {0.0386711} {0.0278364} {0.0179948} {0.00919032} {0.00144176} {–0.00525428} {–0.0109202} {–0.0155945} {–0.0193286}

This detailed study using a time increment of 0.1 reaffirms that the overshoot occurs at t = 2.6 and X1 has a maximum value equal to 0.28502.

6.3 PROGRAM ODEBVPRK — APPLICATION OF RUNGE-KUTTA METHOD FOR SOLVING BOUNDARY-VALUE PROBLEMS The program OdeBvpRK is designed for numerically solving the linear boundaryvalue problems governed by the ordinary differential equation by superposition of two solutions obtained by application of the Runge-Kutta fourth-order method. To explain the procedure involved, consider the problem of a loaded beam shown in Figure 8. Mathematically, the deflection y(x) satisfies the well-known flexural equation.5 d2y M = dx 2 EI

(1)

where M is the internal moment distribution, E is the Young’s modulus and I is the moment of inertia of the cross section of the beam. For the general case, M, E and I can be function of x. The boundary conditions of this problem are: © 2001 by CRC Press LLC

In[5]: = Y30Trial2 = Y[30]/. Id2 Out[5] = {6.00052} Linear combination of the two trial solutions is now possible by calculating a correct Y (X = 0) which should be equal to the value given in Out[8]. In[6]: = d = Y30Trial2Y30Trial1; a = Y30Trial2/d; b = -Y30Trial1/d; In[7]: = Print[“Alpha = “,a,” Beta = “,b] Out[7] = Alpha = {2.00017} Beta = {–1.00017} In[8]: = YP0 = 0.1*a + 0.2*b Out[8] = {–0.0000174888} Finally, the actual deflection and slope can be obtained by providing the correct set of initial conditions and again applying the Runge-Kutta method. In[9]: = Id = (NDSolve[{Y [X] = = YP[X], YP'[X] = = F[X], Y[0] = = 0, YP[0] = = –0.0000174888}, {Y,YP}, {X,0,30}]) In[10]: = (Do[Print[“X = “, Xv, “ Y = “, Y[Xv]/.Id, “ DY/DX = “, YP[Xv]/.Id], {Xv,0,30}]) Out[10] = X= X= X= X= X= X= X= X= X= X= X= X= X= X= X= © 2001 by CRC Press LLC

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y=

{0.} {–0.0000174645} {–0.0000348041} {–0.0000518936} {–0.0000686077} {–0.0000848222} {–0.000100412} {–0.000115251} {–0.000129215} {–0.00014218} {–0.000154019} {–0.000164608} {–0.000173788} {–0.000181454} {–0.000187493}

DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX =

{–0.0000174888} {–0.0000174262} {–0.0000172387} {–0.0000169262} {–0.0000164887} {–0.0000159262} {–0.0000152387} {–0.0000144262} {–0.0000134887} {–0.0000124262} {–0.0000112387} {–0.00000992619} {–0.00000848869} {–0.00000692619} {–0.00000523869}

X= X= X= X= X= X= X= X= X= X= X= X= X= X= X= X=

15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y= Y=

{–0.000191786} {–0.00019422} {–0.00019468} {–0.000193037} {–0.000189163} {–0.000182912} {–0.000174247} {–0.00016319} {–0.00014959} {–0.000133573} {–0.00011547} {–0.0000951574} {–0.0000725124} {–0.0000483635} {–0.0000233292} {0.0000023099}

DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX = DY/DX =

{–0.00000342619} {–0.00000148869} {0.000000573812} {0.00000276006} {0.00000506839} {0.0000074989} {0.0000100026} {0.0000125002} {0.0000149974} {0.0000174176} {0.0000196352} {0.0000216325} {0.0000233628} {0.0000246913} {0.0000255217} {0.0000258107}

6.4 PROGRAM ODEBVPFD — APPLICATION OF FINITE DIFFERENCE METHOD FOR SOLVING BOUNDARY-VALUE PROBLEMS The program OdeBvpFD is designed for numerically solving boundary-value problems governed by the ordinary differential equation which are to be replaced finite-difference equations. To illustrate the procedure involved, let us consider the problem of an annular membrane which is tightened by a uniform tension T and rigidly mounted along its inner and outer boundaries, R = Ri and R = Ro, respectively. As shown in Figure 11, it is then inflated by application of a uniform pressure p. The deformation of the membrane, Z(R), when its magnitude is small enough not to affect the tension T, can be determined by solving the ordinary differential equation6 Z(R) satisfies Equation 1 is for Ri
and

Z( R o ) = 0

(1)

(2)

If the finite-difference approximation is to be applied for solving Equation 1, we will be seeking not for the expression Z(R) but for the numerical values at a selected stations of R in the interval Ri
This is a fourth-order ordinary differential equation and needs fourth supplementary initial, boundary, or mixed conditions in order to completely solve for the deflection y(x). For the simply supported beam subjected to its own weight, the four boundary conditions are: y=0

and

M = EI

d2y =0 dx 2

at

x=0

and

x=L

(17)

If EI is not a function of x, the central-difference approximation for Equation 16 at x = xk is: y − 4 y k −1 + 6y k − 4 y k +1 + y k +2 w d 4y = at x k  k −2 at x k 4 dx EI ( ∆x)4

(18)

If N in-between stations are selected for determination of the deflections there, then the two boundaries are at k = 0 and k = N + 1. In view of the subscripts k–2 and k + 2 in Equation 18, k can thus only take the values 2 through N–1. For solving y1 through yN, we hence need two more equations which are the two moment conditions in Equation 17. At x0 = 0, the second-order, forward-difference formula can be applied to give M = EI

y − 2 y1 + y 0 d2y at x 0  2 EI at x 0 = 0 2 dx ( ∆x)2

Since y = 0 at the left support x = x0 = 0, the above equation yields −2 y1 + y 2 = 0

(19)

Similarly, at the right support x = xN + 1 = 3L, the second-order, backwarddifference formula can be applied to give M = EI

y − 2 y n + y N −1 d2y at x N +1 N +1 EI at x N +1 = 0 2 dx ( ∆x)2

Since y = 0 at the right support x = xN + 1 = 3L, the above equation yields y N −1 − 2 y N = 0

(20)

Meanwhile, the boundary condition y = 0 at the left support x = x0 = 0 can be substituted into Equation 18 for k = 2 to obtain: −4 y1 + 6y 2 − 4 y3 + y 4 = ( ∆x) © 2001 by CRC Press LLC

4

w at x 2 EI

(21)

MATHEMATICA APPLICATIONS For solving the membrane problem, Mathematica can be applied as follows: In[1]: = Ns = 11; DX = 0.5; c = Table[0,{Ns},{Ns}]; R = Table[–5*DX^2/100,{Ns}]; Print[“R = “,R] Out[1] = R = {–0.0125, –0.0125, –0.0125, –0.0125, –0.0125, –0.0125, –0.0125, –0.0125, –0.0125, –0.0125, –0.0125} In[2]: = (Do[Do[x = 3 + i*DX; If[i = = j, c[[i,j]] = –2;, If[i = = j–1, c[[i,j]] = 1DX/2/x;, If[i = = j + 1, c[[i,j]] = 1 + DX/2/x;, Continue]]], {i,Ns}], {j,Ns}]); Print[“Matrix c = “,c] Out[2] = Matrix c = {{–2, 0.928571, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1.0625,–2, 0.9375, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 1.05556, –2, 0.944444, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1.05, –2, 0.95, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1.04545, –2, 0.954545, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1.04167, –2, 0.958333, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1.03846, –2, 0.961538, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1.03571, –2, 0.964286, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 1.03333, –2, 0.966667, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1.03125, –2, 0.96875}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1.02941, –2}} In[3]: = V = Inverse[c].R Out[3] = {0.0533741, 0.101498, 0.142705, 0.175525, 0.198641, 0.210864, 0.211107, 0.198368, 0.171723, 0.13031, 0.0733212} These results are in agreement with those obtained by the MATLAB application. The loaded beam problem also can be treated in a similar manner as follows: In[1]: = c = Table[0,{9},{9}]; c[[1,1]] = –2; c[[1,2]] = 1; c[[9,9]] = –2; c[[9,8]] = 1; In[2]: = (Do[Do[If[i = = j,c[[i,j]] = 6;, If[(j = = i–1)||(j = = i + 1), c[[i,j]] = –4;, If[(j = = i + 2)||(j = = i–2), c[[i,j]] = 1;, Continue]]], {i,2,8}],{j,9}]); Print[“Matrix c = “,c] Out[2] = Matrix c = {{–2, 1, 0, 0, 0, 0, 0, 0, 0}, {–4, 6, –4, 1, 0, 0, 0, 0, 0}, {1, –4, 6, –4, 1, 0, 0, 0, 0}, {0, 1, –4, 6, –4, 1, 0, 0, 0}, {0, 0, 1, –4, 6, –4, 1, 0, 0}, {0, 0, 0, 1, –4, 6, –4, 1, 0}, © 2001 by CRC Press LLC

{0, 0, 0, 0, 1, –4, 6, –4, 1}, {0, 0, 0, 0, 0, 1, –4, 6, –4}, {0, 0, 0, 0, 0, 0, 0, 1, –2}} In[3]: = R = Table[0,{9}]; WE = –2; WM = –2; EIE = 1; EIM = 1; L = 100; DX = 30; In[4]: = (Do[X = i*DX; If[XL, If[X<2*L, R[[i]] = DX^4*WM/EIM;, If[X = = 2*L, R[[i]] = DX^4*(WE + WM)/(EIE + EIM);, IF[X>2*L, R[[i]] = DX^4*WE/EIE;, Continue]]]]]], {i,2,8}]); Print[“R = “,R] Out[4] = R = {0, –1620000, –1620000, –1620000, –1620000, –1620000, –1620000, –1620000, 0} In[5]: = V = Inverse[c].R Out[5] = {–34020000, –68040000, –96390000, –115020000, –121500000, –115020000, –96390000, –68040000, –34020000} Again, the results are in agreement with those obtained by the MATLAB application.

6.5 PROBLEMS RUNGEKUT 1. The differential equation of motion of a spring-and-mass system is d2x/dt2 + ω2x = 0 where ω2 = k/m, m is the mass and k is the spring constant. If the weight is 5 lb., g = 32.2 ft/sec2, k = 1.5 lb/in, and initially the displacement x = 2 in. and velocity dx/dt = 0, use the 4th-order Runge-Kutta method and a step-size of 0.05 sec. to manually calculate the values of x and dx/dt at t = 0.05 sec. Note that the given second-order ordinary differential equations should first be converted into two first-order differential equations. 2. Initially, the two functions x(t) and y(t) have values 1 and –1, respectively. That is x(t = 0) = 1 and y(t = 0) = –1. For t>0, they satisfy the differential equations dx/dt = 5x–2y + 2t and dy/dt = x2–0.25sin2t. Use Runge-Kutta classic fourth-order method and a time increment of 0.01 second to calculate x(t = 0.01) and y(t = 0.01). 3. In the following two equations, the terms dx/dt and dy/dt both appear. 2dx/dt –3x + 5dy/dt –7y = .5t + 1 –3dx/dt -x – 4dy/dt + 9y = .2e –3t © 2001 by CRC Press LLC

Separate them by treating them as unknowns and solve them by simple substitution or Cramer’s rule. The resulting equations can be expressed in the forms of dx/dt = F1(t,x,y) and dy/dt = F2(t,x,y). Carry out by manual computation using Runge-Kutta method to obtain the x and y values when t = 0.1 if the time increment is 0.1 and the initial conditions are x(t = 0) = 1 and y(t = 0) = –2. 4. The distributed loads on the beam shown in Figure 3 can be described as w = –1 N/cm, for 0<x<10 cm; w = 2.2x N/cm, for 10<x<20 cm; w = 0, for 20<x<40 cm. Meanwhile, the bending moment applied at x = 30 cm can be described as Ma = 0, for 0<x<30 cm and Ma = –3 N-cm, for 30<x<40 cm. By introducing new variables t, and x1 through x4 so that t = x, x1 = y, x2 = dy/dx, x3 = d2y/dx2, and x4 = d3y/dx3, the problem of finding the deflection y(x) of the beam can be formulated (see the reference cited in footnote) as: dx1/dt = F1(t,x1,x2,x3,x4) = x2,

x1(t = 0) = 0

dx2/dt = F2(t,x1,x2,x3,x4) = (x3 + Ma)/EI,

x2(t = 0) = 0

dx3/dt = F3(t,x1,x2,x3,x4) = x4,

x3(t = 0) = –1121/3

dx4/dt = F4(t,x1,x2,x3,x4) = w,

x4(t = 0) = 24

with Ma and w being the applied bending moment and distributed loads, respectively. The initial conditions specified above are all at the left end of the beam which is built into the wall and for the deflection (x1), slope (x2), bending moment (x3), and shearing force (x4), respectively. Apply the program RungeKut by using EI = 2x105 N/cm2 and various stepsizes to tabulate the results and errors similar to that given in the text. 5. Apply the fourth-order Runge-Kutta method to find the values of x1 and x2 at the time t = 0.2 second using a time increment of 0.1 second based on the following governing equations: dx1 = 4 − 5x1 + 7x 2 − 2 t dt

and

dx 2 = −3x1 + x1x 2 + 6e − t dt

At t = 0.1 second, x1 = –1 and x2 = 1. 6. Use different stepsizes to calculate y values at x = 0.1, 0.2, and 0.3 by application of the program RungeKut for the initial-value problem dy/dx = x2–y, y(x = 0) = 1. The analytical solution is y = 2–2x + x2e-x, by which the exact solutions can be easily computed to be y(0.1) = 0.90516, y(0.2) = 0.82127, and y(0.3) = 0.74918. Determine the stepsize which will lead to a Runge-Kutta numerical solutions of y(0.1), y(0.2), and y(0.3) accurate to five decimal figures. 7. For the loaded beam shown in Figure 13, the deflection y(x) is to be determined by solving Equation 21. Let the stiffness EI be equal to 2x107 N-cm2 and it can be shown that the bending moment M can be described by the equations: © 2001 by CRC Press LLC

dD = r3C − r4 D dt dE = r4 D dt

10.

11.

12.

13.

14.

This FUNCTION is to be used by the subroutine RKN in application of the fourth-order Runge-Kutta method. The functions x(t) and y(t) satisfy the differential equations d2x/dt2 + 3dx/dt + 5dy/dt–7e–9t + 9sin2t = 0 and 2dx/dt–4dy/dt + 6x–8y + 10t–12 = 0. Convert the above two equations into the standard form dxi/dt = fi(t; x1,x2,x3;constants) for i = 1,2,3 where x1 = x, x2 = y, and x3 = dx/dt. Give the expressions for f1, f2, and f3 in terms of t, x1, x2, and x3 so that the Runge-Kutta method can be applied. Apply the fourth-order Runge-Kutta method to find the values of y and z at x = 0.35 if at x = 0.3, y = 1 and z = 2 respectively and they satisfy thedifferential equations dy/dx = xy + z and dz/dx = yz + x. Use a stepsize of x = 0.5 and show all details of how the Runge-Kutta parameters are calculated. The deflection y of the load beam shown below satisfies the ordinary differential equation EI(d2y/dx2) = M where the Young’s modulus E = 2x108 N/m2, moment of inertia I = 4.5x10–8 m4 and the internal bending moment, in N-m, has been derived in terms of x as M(x) = 200x–30 for 0≤x≤.1 m and M(x) = 100x–20 for .1≤x≤.2 m. (1) Using an increment of x = 0.01 m, standardize the above problem into a system of two firstorder ordinary differential equations dxi/dx = fi(x;x1,x2;constants) for i = 1,2 where x1 = y and x2 = dy/dx (slope). (2) Write a FUNCTION F(…) needed in SUBROUTINE RKN which we have discussed in class for using the fourth-order Runge-Kutta method, based on the result of Step (1) and also the M(x) equations. (3) Calculate the eight RungeKutta parameters and then the value of y and dy/dx at x = 0.01 m. Convert the following differential equation into a set of two first-order differential equations so that the fourth-order Runge-Kutta method can be applied: d2x/dt2 + 4dx/dt + 3x = 4e-t. If at t = 0, x = 0 and dx/dt = 2, use a time increment oft = 0.1, compute the x and dx/dt values at t = 0.1 based on the fourth-order Runge-Kutta method. The forced swaying motion of a three-story building can be simulated as a system of three lumped masses mi connected by springs with stiffnesses ki and subjected to forces fi(t) for i = 1,2,3 as shown in Figure 14. Here, the dampingcharacteristics are not considered but could be incorporated. Derive the governing differential equations for the and then convert them into a system of 6 first-order differential equations so that the programs RungeKut and ode45 can be applied to find the histories of displacements, xi(t), and velocities vi(t) = dxi(t)/dt. Solve a numerical case of mi = 2i N-sec2/cm, ki = 3i N/cm, and fi(t) = (2i–3)sin(2i–1)t N, xi(t = 0) = 0 and vi(t = 0) = 0 for 0
4. Roundoff errors in the Gaussian elimination steps begin to affect the accuracy of the computed values of the deflection for Problem 3 when N = 119. Change the program OdeBvpFD into double precision arithmetics and rerun the case N = 119 and compare the computed ymax to that of analytical solution. 5. Make necessary changes in the FORTRAN version of program OdeBvpFD to solve Problems 3 and 4. 6. For the second sample problem (deflection of beam, Figure 2), change the distributed loads to wm = 2 and we = 1, and the rigidities to EIm = 1 and EIe = 0.5 to recalculate the maximum deflection ymax. 7. Show that for the beam deflection problem shown in Figure 2 when Equation 13 is approximated by use of second-order, central difference and by incorporating the boundary conditions y = 0 at x = 0 and x = 3L, it will lead to the solution of the matrix equation [C]{Y} = {R} where the elements of [C] and {R} denoted as ci,j and ri, respectively can be calculated by the formulas: c i ,i

= −2,

ci,i +1 = ci +1,i = 1, ci, j

= 0,

for i = 1, 2,…, N for i = 1, 2,…, N − 1

(a)

elsewhere

and ri = ( ∆x)

8.

9.

10.

11.

2

M at x i EI

where N is the number of stations between the two supports and x is the stepsize equal to 3L/(N + 1). For Figure 2, if the uniformly distributed loads for the middle and ending portions are designated as wm and we, respectively, derive the expressions for the internal bending moments in the three portions of the beam, 0≤x≤L, L≤x≤2L, and 2L≤x≤3L. Prepare subprogram FUNCTIONS CIJ and RI for Problem 8 and find the deflection vector {Y} by use of either FORTRAN or QuickBASIC version of the program OdeBvpFD. Select appropriate values for the number of stations N so that the results obtained by this second-order approach can be compared to those by the fourth-order approach. Use the central finite-difference method and an increment of x = 1 to find the y values at x = 1 and x = 2 when y is governed by the equation d2y/dx2 + 3dy/dx – y = 2x – 3 and satisfies the boundary conditions y = 0 at x = 0 and x = 3. Use the central finite-difference method and an increment of x = 1 to find the y values at x = 1 and x = 2 when y is governed by the equation d2y/dx2 + 3y = x–1 and satisfies the boundary conditions y = 0 at x = 0 and x = 3.

© 2001 by CRC Press LLC

(b)

12. It is known that u = 0 at r = 2 and r = 5 and that for 2
6.6 REFERENCES 1. C. R. Wylie, Jr., Advanced Engineering Mathematics, McGraw-Hill, New York, 1960, Chapter 6. 2. J. Water, “Methods of Numerical Integration Applied to System Having Trivial Function Evaluation,” ACM Communication, Vol. 9, 1966, p. 293. 3. A. Higdon et al., Mechanics of Materials, John Wiley & Sons, New York, 1985, Chapter 7. 4. Y. C. Pao, Elements of Computed-Aided Design and Manufacturing, CAD/CAM, John Wiley & Sons, New York, 1984. 5. A. Higdon, E. H. Ohlsen, W. B. Stiles, J. A. Weese, and W. F. Riley, Mechanics of Materials, 4th Edition, John Wiley & Sons, New York, 1985. 6. W. Jaunzemis, Continuum Mechanics, MacMillan, New York, 1967, p. 365. 7. S. Timoshenko and D. H. Young, Elements of Strength of Materials, 5th Edition, Van Nostrand Reinhold Co., New York, 1968. 8. W. Jaunzemis, Continuum Mechanics, MacMillan, New York, 1967, p. 365. 9. J. L. Meriam and L. G. Kraige, Engineering Mechanics, Volume One: Statics, Third Edition, John Wiley & Sons, Inc., New York, 1992.

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the common frequency ω is introduced for the free vibration it becomes a standard eigenvalue and eigenvector problem described by the matrix [A]. Hence, the question becomes how to find the eigenvalues and their associated eigenvector of a prescribed matrix. The methods of solution are to be discussed in Sections 7.3 and 7.4 where the programs CharacEq and EigenVec are introduced. In Section 7.5, an iterative method for finding the eigenvector when an eigenvalue of a matrix is provided and program EigenvIt also will be presented. Prior to these discussions, in the next section we will first concern with how the matrices connected with the buckling and vibration problems are to be derived and demonstrate in advance how the programs CharacEq, Bairstow, EigenVec, and EigenvIt are to be employed for obtaining the eigenvalues and eigenvectors of these matrices.

7.2 PROGRAMS EigenODE.Stb AND EigenODE.Vib — FOR SOLVING STABILITY AND VIBRATION PROBLEMS In order to obtain numerical solution of the buckling load and shape of the rod shown in Figure 1 in Section 7.1, the central-difference method introduced in Chapter 4 can be applied to approximate the second derivative term appearing in Equation 1 there. At a typical location along the rod, say x = xj, Equation 1 can be approximated as: Py j d 2 y y j−1 − 2 y j + y j+1 =˙ =− 2 2 dx h (EI) j

(1)

where (EI)j is the rigidity of the rod at xj and yj ≡ y(x = xj) etc. This approach requires that the rod be investigated at N stations between the two supports which are labeled as x0 and xN + 1. These stations are equally spaced so that the increment of x (stepsize h) is simply h = x = L/(N + 1). As a result of such arrangement, the boundary conditions previously defined in Equation 2 now become: y 0 = y N +1 = 0

(2)

By writing out the equation for the first and last in-between stations, i.e., j = 1 and j = N, based on Equation 1 and the boundary conditions (2), the two simplified equations are, respectively:  h2P  y +y =0 −2 + (EI)1  1 2 

(3)

 h2P  y N −1 + −2 + y = 0 (EI)N  N 

(4)

and

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MATHEMATICA APPLICATIONS For the buckling problem, Mathematica can be applied as follows: In[1]: = Ns = 5; EI = 1.; L = 1.; H = L/(Ns + 1); In[2]: = (Print[“Number of Station = “, Ns, “ EI = “, EI, “ Length = “, L, “Delta L = “, H) Out[2]: = Number of Station = 5 EI = 1. Length = 1. Delta L = 0.166667 In[3]: = (Do[Do[If[i == j, M[[i,j]] = 2.*EI/H^2, If[i == (j + 1)||i = = (j–1), M[[i,j]] = EI/H^2, 0]], {i,Ns}],{j,Ns}]); MatrixForm[M] Out[3]//MatrixForm: = 72. 36. 0. 0. 0.

36. 72. 36. 0. 0.

0. 36. 72. 36. 0.

0. 0. 36. 72. 36.

0. 0. 0. 36. 72.

In the next section, we will show how the characteristic equation for the above derived matrix [M] can be determined by application of Mathematica and subsequently how the eigenvalues and eigenvectors are to be obtained.

7.3 PROGRAM CHARACEQ — DERIVATION OF CHARACTERISTIC EQUATION OF A SPECIFIED SQUARE MATRIX The program CharacEq is designed to generate the coefficients of the characteristic equation of an interactively specified square matrix by use of the Feddeev-Leverrier method. Such a characteristic equation is needed in the stability, vibration, and other so-called eigenvalue problems.3 Readers interested in these problems should also refer to the discussions on the programs EigenODE and EigenVec. The former discusses how the square matrix is to be generated by finite-difference approximation of ordinary differential equation. The latter program delineates how the eigenvectors are to be found by a modified Gaussian elimination method for each eigenvalue and how the eigenvalues are to be solved from the characteristic equation by the program Bairstow. Here for derivation of the characteristic equation, let us denote the specified square matrix be [A] and its elements be ai,j for i,j = 1,2,…,n with n being the order of [A]. The Feddeev-Leverrier method first express the characteristic equation of [A] as:

(−1)n (λn − p1λn −1 − p2 λn −2 − … − p n −1λ − p n ) = 0 © 2001 by CRC Press LLC

(1)

Output[1] = m = {{0,2,3), {–10,–1,2}, {–2,4,7}} Notice that the elements in each row are separated by comma and enclosed by a pair of braces, and rows are separated also by comma. Next, we derive the characteristic equation of the matrix m. Input[2]: = Det[m — x IdentityMatrix[3]] Output[2] = 6 – 11 X + 6 X2 — X3 Input[3]: = m = {{1,2,3), {–10,0,2}, {–2,4,8}} Output[3] = m = {{1,2,3), {–10,0,2}, {–2,4,8}} Input[4]: = Det[m — x IdentityMatrix[3]] Output[4] = 24 – 26 X + 9 X2 – X3 We may proceed to solve the characteristic roots as follows: Input[5]: = NSolve[24–26x + 9x^2x^3 = = 0,x] Output[5] = {{x -> 2.}, {x -> –3.}, {x -> 4.}} Again, the polynomial can be plotted with: Input[6]: = Plot[x^3–9x^2 + 26x–24, {x,1,5}, Frame->True}, AspectRatio->1]

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In[1]: = a = {{3,0,2},{0,5,0},{2,0,3}}; MatrixForm[a] Out[1]//MatrixForm: = 3 0 2

0 5 0

2 0 3

In[2]: = Eigenvalues[a] Out[2]: = {1, 5, 5} In[3]: = Eigenvectors[a] Out[3]: = {–1, 0, 1}, {1, 0, 1}, {0, 1, 0}} Notice that the computed eigenvectors are not normalized. As another example, consider the matrix M generated in the program EigenODE for the buckling problem when the number of stations is equal to 5. To obtain the eigenvalues, the interactive application of Mathematica goes as: In[4]: = Eigenvalues[M] Out[4]: = {134.354, 108., 72., 36., 9.64617} Notice that the smallest eigenvalue is equal to 9.64617 which predicts the lowest buckling load. Since the exact solution is 9.8696, this further indicates that by continuously increasing the number of stations the smallest eigenvalue in magnitude will eventually converge to the expected value.

PRINCIPAL STRESSES

AND

PLANES

As another example of solving the eigenvalues and eigenvectors, consider the problem of determining the principal stresses at a point within a two-dimensional body which is subjected to in-plane loadings. If the normal stresses (x and σy) and shear stresses (xy = yx), Figure 5, at that point are known, it is a common practice to graphically determine the principal stresses and principal planes, on which the principal stresses act by use of Mohr’s circle.4 But, here we demonstrate how the principal stresses and principal planes can be solved as the eigenvalues and eigenvectors, respectively, of a matrix [A] constructed using the values of x, y, and xy as follows:

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cos[90°– min] = –0.9732). The two principal planes are perpendicular to each other. This can also be proven by taking the dot product of the two normalized eigenvectors: (–0.9732i + 0.2298j)•(–0.2298i- 0.9732j) = 0. Similar observation can be made from the results for Problem (b). The principal stresses are equal to 16.9085, 34.6906, and 58.4009 N/cm2 and they on the planes having outward normal vectors n1 = 0.8632i + 0.4921j + 0.1192k, n2 = 0.3277i – 0.7218j + 0.6096k, and n3 = –0.3860i + 0.4867j + 0.7837k, respectively. It is easy to prove that these principal planes are indeed orthogonal by showing that n1•n2 = n2•n3 = n1•n3 = 0.

QUADRATIC FORMS

AND

CANONICAL TRANSFORMATION

Another interesting application of the procedure involved in solving eigenvalues and eigenvectors of a square matrix is the canonical transformation of quadratic forms6 for Consider a surface described by the equation: 25x 2 + 36y 2 + 49z 2 − 24 xy − 18xz + 16yz = 100

(4)

The left-hand side is called a quadratic form in x, y, and z. The surface is an ellipsoid but it is difficult to make out what are the values of its major and minor axes, for which a transformation of the coordinate system is necessary for changing the quadratic form into a canonical one. By canonical form, it means that only the squared terms should remain. To find what transformation is needed, we first write Equation 4 in matrix form as:

{V}T [A]{V} = 100

(5)

where: x  {V} = y  z 

and

 25 [A] = −12  −9

−12 36 8

−9 8  49 

(6,7)

It can be shown6 that if we find the eigenvalues and eigenvectors of [A] which in fact are already available in the answer to the previously discussed Problem (b), the coordinate system x-y-z can be transformed into another set of x -y -z system by using the so-called normalized modal matrix, [Q], formed with the normalized eigenvectors as its columns. That is: x′  .8623 V = { ′} y′ = .4921  z′  .1192 © 2001 by CRC Press LLC

.3277 −.7218 .6096

−.3860  .4867  .7837 

x  y  = Q {V}   [ ]  z 

(8)

should be terminated. To obtain the largest eigenvalue in magnitude, we proceed directly with a given matrix as follows: Input[1]: = A = {{2,0,3},{0,5,0},{3,0,2}}; V = {1,0,0}; I = 0; Input[2]: = While[i<8, I = I + 1; VN = A.V; Lambda = Sqrt[VN.VN]; Print["Iteration # ",I," ","Lambda = ",N[Lambda], " {V} =",V = N[VN/Lambda]]] Output[2]: = Iteration Iteration Iteration Iteration Iteration Iteration Iteration Iteration

# # # # # # # #

1 2 3 4 5 6 7 8

Lambda = Lambda = Lambda = Lambda = Lambda = Lambda = Lambda = Lambda =

3.60555 {V} = {0.5547, 0, 0.83205} 4.90682 {V} = {0.734803, 0, 0.67828} 4.99616 {V} = {0.701427, 0, 0.712741} 4.99985 {V} = {0.708237, 0, 0.709575} 4.99999 {V} = {0.70688, 0, 0.707333} 5. {V} = {0.707152, 0, 0.707062} 5. {V} = {0.707098, 0, 0.707016} 5. {V} = {0.707109, 0, 0.707105}

Notice that the testing condition is whether the running iteration counter I is less than 8 or not. This setup enables up to eight iterations to be conducted. The function N in Input[2] requests the value of the variable inside the brackets to be given in numerical form. For example, when the value of Lambda is displayed as sqrt[4], it will be displayed as 2.00000 if the input is N[Lambda]. VA·VB is the dot product of VA and VB. In Input[2] some sample printouts of the character strings specified inside a pair of parentheses are also demonstrated. For iterating the smallest eigenvalue in magnitude, we work on the inverse of this given matrix as follows: Input[3]: = Ainv = Inverse[A]; V = {1,0,0}; I = 0; Input[4]: = While[i<8, I = I + 1; VN = Ainv.V; Lambda = Sqrt[VN.VN]; Print["Iteration # ",I," ","Lambda = ",N[Lambda], " {V} = ",V = N[VN/Lambda]]] Output[4]: = Iteration Iteration Iteration Iteration Iteration Iteration Iteration Iteration

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# # # # # # # #

1 2 3 4 5 6 7 8

Lambda = Lambda = Lambda = Lambda = Lambda = Lambda = Lambda = Lambda =

0.72111 {V} = {–0.5547, 0, 0.83205} 0.981365 {V} = {0.734803, 0, –0.67828} 0.999233 {V} = {–0.701427, 0, 0.712741} 0.999969 {V} = {0.708237, –0, 0.709575} 0.999999 {V} = {–0.70688, 0, 0.707333} 1. {V} = {0.707152, 0, –0.707062} 1. {V} = {–0.707098, 0, 0.707016} 1. {V} = {0.707109, 0, –0.707105}

Inverse is a Mathematica function which inverts a specified matrix. The above printout of eight iterations shows that {V} continues to change its sign. This is an indication that the eigenvalue carries a minus sign.

7.6 PROBLEMS PROGRAMS EIGENODE.STB

AND

EIGENODE.VIB

1. Apply the program EigenODE.Stb for the cases N = 6, 7, and 8 to obtain the coefficient matrix [C] in the standard eigenvalue problem of ([C]λ[I]){Y} = {0}. 2. Apply the program CharacEq to obtain the characteristic equations of order 6, 7, and 8 for the matrices [C] derived in Problem 1. 3. Apply the program Bairstow to find the roots for the characteristic equationsderived in Problem 2. If necessary, change this program to allow interactive input of the u and v values for the guessing quadratic factor 2 + u  + v. This enhancement will be helpful if the iteration fails to converge. 4. The program EigenODE.Vib has been arranged for solving the general problem of having N masses, m1–mN, in series connected by N + 1 springs with stiffnesses k1–kN + 1. Apply it for the case when the three masses shown in Figure 1 are connected by four springs with the fourth spring attached to the ground. Use m1 = m2 = m3 = 1 and k1 = k2 = k3 = k4 = 10. 5. Apply the program CharacEq to find the characteristic equation for the vibration problem described in Problem 4. 6. Apply the program MatxInvD to invert the matrix obtained in Problem 4 and then apply the program EigenvIt to iteratively determine its smallest eigenvalue in magnitude and associated eigenvector. 7. Extend the vibrating system described in Problem 4 to four masses and five springs and then implement the application of the programs MatxInvD and EigenvIt as described in Problems 5 and 6, respectively. 8. Apply the programs CharacEq, Bairstow, and EigenVec to find the characteristic equation, eigenvalues, and associated eigenvectors for the matrix derived in Problem 4, respectively. 9. Same as Problem 8 except for a four masses and five springs system. 10. An approximate analysis of a three-story building is described in Problem 7 in the program EigenvIt. Derive the governing differential equations for the swaymotions xi for i = 1,2,3 and then show that the stiffness matrix [K] and mass matrix [M] are indeed as those given there.

CHARACEQ 1. Apply Feddeev-Leverrier method to find the characteristic equation of the matrix:

© 2001 by CRC Press LLC

1 4   7

2 5 8

3 6 9

2. Apply Feddeev-Leverrier method to find the characteristic equation of the matrix:  5 −10   −2

0 6 0

1 0  7

3. Apply Feddeev-Leverrier method to find the characteristic equation of the matrix:  2 −10   −2

2 1 4

3 2  9

4. Apply the program CharacEq for solving Problems 1 to 3. 5. Apply Feddeev-Leverrier method to find the characteristic equation of the matrix: 5 2  0

0 6 3

1 0  7

6. Apply poly.m of MATLAB to Problems 1 to 3 and 5. 7. Find the roots of the polynomials found in Problem 6 by application of roots.m of MATLAB. 8. Apply plot.m of MATLAB for the polynomials obtained in Problem 6. 9. Apply the function det of Mathematica to derive the characteristic equation for the matrix given in Problem 1. 10. Apply the function det of Mathematica to derive the characteristic equation for the matrix given in Problem 2. 11. Apply the function det of Mathematica to derive the characteristic equation for the matrix given in Problem 3. 12. Apply the function det of Mathematica to derive the characteristic equation for the matrix given in Problem 5.

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EIGENVEC 1. Run the QuickBASIC version of the program EigenVec for the sample case used in the FORTRAN version. 2. Apply the program EigenVec to find the eigenvector corresponding to the eigenvalue equal to 4.41421 for the matrix: 2 0  1

0 3 0

3 0  4

3. Apply the program CharacEq to find the characteristic equation for matrix: 1 4   7

3 6  10 

2 5 8

and then apply the program Bairstow to find the eigenvalues. Finally, apply the program EigenVec to find the eigenvectors. 4. Apply the program CharacEq to find the characteristic equation for the matrix:  5 −10   −2

0 6 0

1 0  7

and then apply the program Bairstow to find the eigenvalues. Finally, apply the program EigenVec to find the eigenvectors. 5. Apply the program CharacEq to find the characteristic equation for the matrix:  2 −10   −2

2 1 4

3 2  9

and then apply the program Bairstow to find the eigenvalues. Finally, apply the program EigenVec to find the eigenvectors. 6. The eigenvalues for the following matrix have been found to be equal to 9.3726, 32 and 54.627:

© 2001 by CRC Press LLC

32 16   0

0 16  32 

16 32 16

Find the associated eigenvector by applying the program EigenVec. 7. The eigenvalues of the following matrix have been found to be equal to 9.5492, 34.549, 64.451, and 90.451: 50 25  0  0

25 50 25 0

0 0  25  50 

0 25 50 25

Find the associated eigenvector by applying the program EigenVec. 8. Find the eigenvalue and associated eigenvector of the matrix: 76 36  0  0  0

36 72 36 0 0

0 36 72 36 0

0 0  0  36  72 

0 0 36 72 36

9. Swaying motion of a three-story building is described in Problem 7 in the program EigenvIt. Use the data there to form the matrix [A] which is equal to [K]–1 [M]. Apply the programs CharacEq and Bairstow to find all three eigenvalues and then apply the program EigenVec to find the associated eigenvectors. 10. Apply the function eig.m of MATLAB to find all eigenvalues of the matrices given in Problems 2 to 8. 11. Apply the functions eigenvalues and eigenvectors of Mathematica to find all eigenvalues of the matrices given in Problems 2 to 8.

EIGENVIT 1. Using an initial, guessed eigenvector {V} = [1 0 0]T, perform four iterative steps to find the largest eigenvalue in magnitude and its associated normalized vector of the matrix:  2 [A] = −10  −2 © 2001 by CRC Press LLC

2 1 4

3 2  9

2. Using an initial, guessed eigenvector {V} = [1 0 0]T, perform four iterative steps to find the largest eigenvalue in magnitude and its associated normalized vector of the matrix: 2 [A] = 0 1

1 0  4

0 3 0

3. Apply the program EigenvIt to find the largest eigenvalue in magnitude and its associated normalized eigenvector of the matrix: 5 1

[A] = 

4.

5. 6.

7.

3 3

Next, apply the program MatxInvD to find the inverse of [A] which is to be entered as input for program EigenvIt to iterate the smallest eigenvalue in magnitude and its associated normalized eigenvector for [A]. Compare the results with the analytical solution of smallest = 2 and largest = 6. Apply the program MatxInvD to find the inverse of the matrix [A] given in Problem 1 and then apply the program EigenvIt to find the smallest eigenvalue in magnitude and its associated normalized eigenvector of [A]. For checking the values of smallest obtained here and largest obtained in Problem 1, derive the characteristic equation of [A] by use of the program CharacEq and solve it by application of the program Bairstow. Same as Problem 4 but for the matrix [A] given in Problem 2. Apply poly.m, roots.m, polyval.m, plot.m, and xlabel and ylabel to obtain a plot of the characteristic equation of the matrix [A] given in Problem 1, shown in Figure 7, to know the approximate locations of the characteristic roots. For a 3-floor building as sketched in the left side of Figure 8, an approximate calculation of its natural frequencies can be attempted by using a lumped approach which represents each floor with a mass and the stiffnesses of the supporting columns by a spring as shown in the right side of Figure 8. If the swaying motion of the floors are expressed as xi = Xisint for i = 1,2,3 where  is the natural frequency and Xi are the amplitudes, it can be shown that  and {X} = [X1 X2 X3]T satisfy the matrix equation [K]{X} = 2[M]{X}, in which the mass matrix [M] and stiffness matrix [K] are formed by the masses and spring constants as follows: m1 [M] =  0  0

© 2001 by CRC Press LLC

0 m2 0

0  0 m 3 

and

 k1 [K] = − k 2  0

−k2 k1 + k 2 −k3

0   −k3  k 2 + k 3 

8.

9. 10.

11. 12.

To find the lowest natural frequency min, the program EigenvIt can be applied to obtain the max from the matrix equation [A]{X} = {X} where the matrix [A] is equal to [K]–1[M] and = 2. min is equal to 1/max. Determine the numeric value of min for the case when m1 = 8x105, m2 = 9x105, and m3 = 1x106 all in N-sec2/m, and k1 = 3x108, k2 = 4x108, and k3 = 5x108 all in N/m. Referring to Figure 2 in the program EigenVec, iteratively determine the maximum and minimum principal stresses and their associated principal planes at a point where the two-dimensional normal and shear stresses are x = 50, y = –30, and xy = yx = –20 all in N/cm2. Compare the results with those obtained in the program EigenVec. Same as Problem 8, except for a three-dimensional case of x = 25, σy = 36, z = 49, xy = yx = –12, yz = zy = 8, and zx = xz = –9, all in N/cm2. Apply MATLAB to invert the matrix [A] given in Problem 1 and then apply EigenvIt.m to iterate the eigenvalue which is the smallest in magnitude and also the associated eigenvector. Same as Problem 10 but for the matrix [A] given in Problem 2. Apply Mathematica to solve Problems 10 and 11.

7.7 REFERENCES 1. W. F. Riley and L. Zachary, Introduction to Mechanics of Materials, Wiley & Sons, Inc., New York, 1989. 2. K. N. Tong, Theory of Mechanical Vibration, Wiley & Sons, Inc., New York, 1960. 3. Y. C. Pao, “A General Program for Computer Plotting of Mohr’s Circle,” Computers and Structures, V. 2, 1972, pp. 625–635. This paper discusses various sources of how eigenvalue problems are formed and also methods of analytical, computational, and graphical solutions. 4. Y. C. Pao, “A General Program for Computer Plotting of Mohr’s Circle,” (for twodimensional cases), Computers and Structures, V. 2, 1972, pp. 625–635. 5. F. B. Seely and J. O. Smith, Advanced Mechanics of Materials, Second Edition, John Wiley, New York, 1957, pp. 59–64. 6. F. B. Hilebrand, Methods of Applied Mathematics, Prentice-Hall, Englewood Cliffs, NJ, 1960. 7. S. Perlis, Theory of Matrices, Addison-Wesley Publishing Company, Reading, MA, 1952.

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8

Partial Differential Equations

8.1 INTRODUCTION Different engineering disciplines solve different types of problems in their respective fields. For mechanical engineers, they may need to solve the temperature change within a solid when it is heated by the interior heat sources or due to a rise or decrease of its boundary temperatures. For electrical engineers, they may need to find the voltages at all circuit joints of a computer chip board. Temperature and voltage are the variables in their respective fields. Hence, they are called field variables. It is easy to understand that the value of the field variable is spacedependent and time-dependent. That is to say, that we are interested to know the spatial and temporal changes of the field variable. Let us denote the field variable as . and let the independent variables be xi which could be the time t, or, the space coordinates as x, y, and z. In order not to overly complicate the discussion, we introduce the general two-dimensional partial differential equation which governs the field variable in the form of: A(x1, x 2 )

∂2 φ ∂2 φ ∂2 φ + B(x1, x 2 ) + C(x1, x 2 ) 2 2 ∂x1 ∂x1∂x 2 ∂x 2  ∂ φ ∂φ  , = F x1, x 2 , φ, ∂x1 ∂x 2  

(1)

where the coefficient functions A, B, and C in the general cases are dependent on the variables x1 and x2, and the right-hand-side function F, called forcing function may depend not only on the independent variables x1 and x2 but may also depend on the first derivatives of . There are innumerable of feasible solutions for Equation 1. However, when the initial and/or boundary conditions are specified, only particular solution(s) would then be found appropriate. In this chapter, we will discuss three simple cases when A, B, and C are all constants. The first case is a two-dimensional, steady-state heat conduction problem involving temperature as the field variable and only the spatial distribution of the temperature needs to be determined, Equation 1 is reduced to a parabolic partial differential equation named after Poisson and Laplace when the forcing function F is not equal to, or, equal to zero, respectively. This is a case when the coefficient functions in Equation 1 are related by the condition B2–4AC<0. The second case is a one-dimensional, transient heat conduction problem. Again, the field variable is the temperature which is changing along the longitudinal x-axis © 2001 by CRC Press LLC

of a straight rod and also in time. That is, x1 becomes x and x2 become the time t. Equation 1 is reduced to an elliptical partial differential equation. This is a case when B2–4AC = 0. The third case is the study of the vibration of a tightened string. The field variable is the lateral deflection of this string whose shape is changing in time. Equation 1 is reduced to a hyperbolic partial differential equation. If x is the longitudinal axis of the string, then same as in the second case, the two independent variables are x and t. This is a case when B2–4AC>0. The reason that these problems are called parabolic, elliptical, and hyperbolic is because their characteristic curves have such geometric features. Readers interested in exploring these features should refer to a textbook on partial differential equations. Details will be presented regarding how the forward, backward, and central differences discussed in Chapter 4 are to be applied for approximating the first and second derivative terms appearing in Equation 1. Repetitive algorithms can be devised to facilitate programming for straight-forward computation of the spatial and temporal changes of the field variable. Numerical examples are provided to illustrate how these changes can be determined by use of either QuickBASIC, FORTRAN, MATLAB, or, Mathematica programs. Although explanation of the procedure for numerical solution of these three types of problems is given only for the simple one- and two-dimensional cases, but its extension to the higher dimension case is straight forward. For example, one may attempt to solve the transient heat conduction problem of a thin plate by having two space variables instead of one space variable for a long rod. The steady-state heat conduction problem of a thin plate can be extended for the case of a three-dimensional solid, and the string vibration problem can be extended to a two-dimensional membrane problem.

8.2 PROGRAM PARABPDE — NUMERICAL SOLUTION OF PARABOLIC PARTIAL DIFFERENTIAL EQUATIONS The program ParabPDE is designed for numerically solving engineering problems governed by parabolic partial differential equation in the form of: ∂φ ∂2φ =a 2 ∂t ∂x

(1)

and  is a function of t and x and satisfies a certain set of supplementary conditions. Equation 1 is called a parabolic partial differential equation. For example,  could be the temperature, T, of a longitudinal rod shown in Figure 1 and the parameter a in Equation 1 could be equal to k/c where k, c, and  are the thermal conductivity, specific heat, and specific weight of the rod, respectively. To make the problem more specific, the rod may have an initial temperature of 0°F throughout and it is completely insulated around its lateral surface and also at its right end. If its left end is to be maintained at 100°F beginning at the time t = 0, then it is of interest to know © 2001 by CRC Press LLC

In[1]: = (n = 11; rk = 0.037; c = 0.212; rho = 168; dt = 1; dx = 0.1; c1 = rk/c/rho*dt/dx^2; nm1 = n–1;) In[2]: = (t = Table[0,{n}]; tn = Table[0,{n}]; t[[1]] = 100; tn[[1]] = 100; tdf = 1; tm = 0; flag1 = 0; In[3]: = (While[flag1 = = 0, Do[tn[[j]] = t[[j]] + c1*(t[[j–1]]–2*t[[j]] + t[[j + 1]]), {j,2,nm1}]; tn[[n]] = t[[n]] + 2*c1*(t[[nm1]]-t[[n]]);tm = tm + dt;flag1 = 1; Do[If[Abs[tn[[I]]-t[[I]]]>tdf, flag1 = 0; Break, Continue],{i,n}]; Do[t[[I]] = tn[[I]],{i,n}]]; Print[“t = “,tm]; Print[N[tn,3]]) Out[3] = t = 24 {100., 65.7, 37.5, 18.5, 7.83, 2.85, 0.892, 0.241, 0.0561, 0.0116, 0.00394} The temperature distribution of the heated rod after 24 seconds is same as obtained by the FORTRAN, QuickBASIC, and MATLAB versions when every component of two consecutive temperature distribution, kept as {t} and {tn}, differ no more than the allowed set value of tdf = 1 degree in In[2]. Any component has a difference exceeding the tdf value will cause flag1 to change from a value of 1 to 0 and the Break command in the second Do loop in In[2] to exit and to continue the iteration. flag1 is created to control the While command which determines when the iteration should be terminated. The N[tn,3] instructs the components of {tn} be printed with 3 significant figures. When tdf is changed to a value of 0.5, Mathematica can again be applied to yield In[4]: = t = (Table[0,{n}]; tn = Table[0,{n}]; t[[1]] = 100; tn[[1]] = 100; tdf = 0.5; tm = 0; flag1 = 0;) In[5]: = (While[flag1 = = 0,Do[tn[[j]] = t[[j]] + c1*(t[[j–1]]–2*t[[j]] + t[[j + 1]]), {j,2,nm1}]; tn[[n]] = t[[n]] + 2*c1*(t[[nm1]]-t[[n]]);tm = tm + dt;flag1 = 1; Do[If[Abs[tn[[I]]-t[[I]]]>tdf, flag1 = 0; Break, Continue],{i,n}]; Do[t[[I]] = tn[[I]],{i,n}]]; Print[“t = “,tm]; Print[N[tn,3]]) Out[5]: = t = 49 {100., 75.5, 53.2, 34.9, 21.3, 12., 6.24, 3., 1.35, 0.617, 0.413} Notice that Mathematica takes only 49 seconds, one second less than that required for the FORTRAN, QuickBASIC, and MATLAB versions. The reason is that Mathematica keeps more significant digits in carrying out all computations. To show more on the effect of changing the tdf value, the following Mathematica runs are provided:

© 2001 by CRC Press LLC

In[6]: = t = (Table[0,{n}]; tn = Table[0,{n}]; t[[1]] = 100; tn[[1]] = 100; tdf = 0.1; tm = 0; flag1 = 0;) In[7]: = (While[flag1 == 0, Do[tn[[j]] = t[[j]] + c1*(t[[j–1]]–2*t[[j]] + t[[j + 1]]), {j,2,nm1}]; tn[[n]] = t[[n]] + 2*c1*(t[[nm1]]-t[[n]]); tm = tm + dt; flag1 = 1; Do[If[Abs[tn[[I]]-t[[I]]]>tdf, flag1 = 0; Break, Continue],{i,n}]; Do[t[[I]] = tn[[I]],{i,n}]]; Print[“t = “,tm]; Print[N[tn,3]]) Out[7]: = t = 462 {100., 93.9, 88., 82.3, 77.1, 72.5, 68.5, 65.3, 63., 61.6, 61.1} In[8]: = t = (Table[0,{n}]; tn = Table[0,{n}]; t[[1]] = 100; tn[[1]] = 100; tdf = 0.05; tm = 0; flag1 = 0;) In[9]: = (While[flag1 == 0, Do[tn[[j]] = t[[j]] + c1*(t[[j–1]]–2*t[[j]] + t[[j + 1]]), {j,2,nm1}]; tn[[n]] = t[[n]] + 2*c1*(t[[nm1]]-t[[n]]); tm = tm + dt; flag1 = 1; Do[If[Abs[tn[[I]]-t[[I]]]>tdf, flag1 = 0; Break, Continue],{i,n}]; Do[t[[I]] = tn[[I]],{i,n}]]; Print[“t = “,tm]; Print[N[tn,3]]) Out[9]: = t = 732 {100., 97., 94., 91.2, 88.5, 86.2, 84.2, 82.6, 81.5, 80.8, 80.5} In[10]: = t = (Table[0,{n}]; tn = Table[0,{n}]; t[[1]] = 100; tn[[1]] = 100; tdf = 0.0001; tm = 0; flag1 = 0; In[11]: = (While[flag1 == 0, Do[tn[[j]] = t[[j]] + c1*(t[[j–1]]–2*t[[j]] + t[[j + 1]]), {j,2,nm1}]; tn[[n]] = t[[n]] + 2*c1*(t[[nm1]]-t[[n]]); tm = tm + dt; flag1 = 1; Do[If[Abs[tn[[I]]-t[[I]]]>tdf, flag1 = 0; Break, Continue],{i,n}]; Do[t[[I]] = tn[[I]],{i,n}]]; Print[“t = “,tm]; Print[N[tn,3]]) Out[11]: = t = 3159 {100., 100., 100., 100., 100., 100., 100., 100., 100., 100., 100.} For tdf = 0.1 and tdf = 0.05, Mathematica continues to take lesser time than the other version; but when tdf = 0.0001, Mathematica needs two additional seconds. The © 2001 by CRC Press LLC

reason is that seven significant figures are required in the last case, rounding may have resulted in earlier termination of the iteration when the FORTRAN, QuickBASIC, and MATLAB versions are employed.

8.3 PROGRAM RELAXATN — SOLVING ELLIPTICAL PARTIAL DIFFERENTIAL EQUATIONS BY RELAXATION METHOD The program Relaxatn is designed for solving engineering problems which are governed by elliptical partial differential equation of the form: ∂2 φ ∂2 φ + = F(x, y) ∂x 2 ∂y 2

(1)

where  is called the field function and F(x,y) is called forcing function. When the steady-state heat conduction of a two-dimensional domain is considered,2 then the field function becomes the temperature distribution, T(x,y), and the forcing function becomes the heat-source function, Q(x,y). If the distribution of T is influenced only by the temperatures at the boundary of the domain, then Q(x,y) = 0 and Equation 1 which is often called a Poisson equation is reduced to a Laplace equation: ∂2 T ∂2 T + =0 ∂x 2 ∂y 2

(2)

The second-order, central-difference formulas (read the program DiffTabl) can be applied to approximate the second derivatives in the above equation at an arbitrary point x = xi and y = yj in the domain as: T − 2 Ti, j + Ti +1, j ∂2 T at x i , y j =˙ i −1, j 2 ∂x ( ∆x)2 and T − 2 Ti, j + Ti, j+1 ∂2 T at x i , y j =˙ i, j−1 2 ∂y (∆y)2 By substituting both of the above equations into Equation 2 and taking equal increments in both x- and y-directions, the reduced equation is, for x = y, Ti, j =

(

1 T + Ti +1, j + Ti, j−1 + Ti, j+1 4 i −1, j

)

(3)

The result is expected when the temperature distribution reaches a steady state because it states that the temperature at any point should be equal to the average of its surrounding temperatures. © 2001 by CRC Press LLC

insulated boundary, mathematically it requires that T/n = 0 there, n being the normal direction. At the vertical boundaries x = xM, we have T/n = T/x = 0 since x is the normal direction. Based on the central difference and considering two increments in the x direction, at a yjth station we can have: − TM −1, j T ∂T =0 at x M , y j =˙ M +1, j 2( ∆x) ∂x

(

)

or TM +1, j = TM −1, j

(5)

Since xM + 1 is below the bottom boundary of the plate shown in Figure 3, there is no need to calculate the temperatures there, however Equation 5 enables the temperatures at the stations along the bottom boundary of the plate TM,j for j = 1 to N to be averaged. Returning to Equation 4, we notice that if Equation 5 is substituted into it, the resulting equation which relates only to three neighboring temperatures is: TM, j =

(

1 2 TM −1, j + TM, j−1 + TM, j+1 4

)

for j = 2, 3,…, N − 1

(6)

Notice that j = 1 and j = N are not covered in Equation 6. These two cases concerning the insulated stations at the left and right bottom corners of the plate will be discussed after we address the two vertical, insulated boundaries y = y1 and y = yN. For the boundaries y = y1, ∂T/∂n becomes ∂T/∂y. Again, we can apply the central difference for double y increments to obtain: T − Ti,2 ∂T at (x1, y1 ) =˙ i,0 =0 2( ∆y) ∂y or Ti,0 = Ti,2

(7)

Thus, the modified Equation 3 for the left insulated boundary is: Ti,1 =

(

1 T + Ti +1,1 + 2 Ti,2 4 i −1,1

)

for

i = 7, 8, 9

In a similar manner, we can derive for the right insulated boundary y = yN −T T ∂T at (x i , y N ) =˙ i,N +1 i,N −1 = 0 2( ∆y) ∂y or © 2001 by CRC Press LLC

(8)

Ti,N +1 = Ti,N −1

(9)

and Ti,N =

(

1 T + Ti +1,N + 2 Ti,N −1 4 i −1,N

)

for i = 8, 9

(10)

Having derived Equations 6, 8, and 10, it is easy to deduce the two special equation for the corner insulated stations to be: TM,1 =

(

1 T + TM −1,1 2 M ,2

)

(11)

and TM,N =

(

1 T + TM −1,N 2 M,N −1

)

(12)

We have derived all equations needed for averaging the temperature at any station of interest including those at the insulated boundaries by utilizing those at its neighboring stations. It suggests that a continuous upgrading process can be developed which assumes that the neighboring temperatures are known. This so-called relaxation method starts with an initial assumed distribution of temperature [T(0)] and continues to use Equations 3 and 6 to 12 until the differences at all locations are small enough. Mathematically, the process terminates when: M

N

i =1

j=1

∑∑ T

( k +1) i, j

− Ti(,kj ) < ε

(13)

where  is a prescribed tolerance of accuracy and k = 0,1,2,… is the number of sweeps in upgrading the temperature distribution. Superscripts (k + 1) and (k) refer to the improved and previous distributions, respectively. The order of sweep will affect how the temperatures should be upgraded. For example, if the temperatures are to be re-averaged from top to bottom and left to right, referring to Figure 1, then Equation 3 is to be modified as: Ti(,kj +1) =

(

1 ( k +1) T + Ti(+k1), j + Ti(,kj−+11) + Ti(,kj+)1 4 i −1, j

)

(14)

Notice that the neighboring temperatures in the row above, i–1, and in the column to the left, j–1, have already been upgraded while those in the row below, i + 1, and in the column to the right, j + 1, are yet to be upgraded. Similar modifications are to be made to Equations 6 to 12 during relaxation. © 2001 by CRC Press LLC

Figure 5b shows the 137th temperature distribution when the interactive MATLAB instructions entered are: >> V = 0:1:50; contour(T,V’), grid, title(‘* After 137 relaxations *’)

Notice that area near the insulated boundaries at the right-lower corner has finally reached a steady-state temperature distribution, i.e., changes of the entire temperature distribution will be insignificant if more relaxations were pursued.

MATHEMATICA APPLICATIONS To apply the relaxation method for finding the steady-state temperature distribution of the heated plate already solved by the FORTRAN, QuickBASIC, and MATLAB versions, here we make use of the Do, If, and While commands of Mathematica to generate similar results through the following interactive operations: In[1]: = t = Table[0,{10},{20}]; eps = 100; nr = 0; d = eps + 1; In[2]: = Do[t[[i,1]] = (I–1)*10,{i,2,6}]; In[3]: = (While[d>eps, d = 0;nr = nr + 1; Do[Do[ts = t[[i,j]]; t[[i,j]] = .25*(t[[I–1,j]] + t[[I + 1,j]] + t[[i,j–1]] + t[[i,j + 1]]); d = Abs[ts-t[[i,j]]] + d;,{j,2,19}],{i,2,6}]; Do[ts = t[[i,1]]; t[[i,1]] = .25*(t[[I–1,1]] + t[[I + 1,1]] + 2*t[[i,2]]); d = Abs[ts-t[[i,1]]] + d; Do[ts = t[[i,j]]; t[[i,j]] = .25*(t[[I–1,j]] + t[[I + 1,j]] + t[[i,j–1]] + t[[i,j + 1]]); d = Abs[ts-t[[i,j]]] + d;,{j,2,19}]; If[i = = 7,Continue, ts = t[[i,20]]; t[[i,20]] = .25*(t[[I–1,20]] + t[[I + 1,20]] + 2*t[[i,19]]); d = Abs[ts-t[[i,20]]] + d;],{i,7,9}]; ts = t[[10,1]]; t[[10,1]] = .5*(t[[9,1]] + t[[10,2]]); d = Abs[ts-t[[10,1]]] + d; Do[ts = t[[10,j]]; t[[10,j]] = .25*(2*t[[9,j]] + t[[10,j–1]] + t[[10,j + 1]]); d = Abs[ts-t[[10,j]]] + d;,{j,2,19}]; ts = t[[10,20]]; t[[10,20]] = .5*(t[[9,20]] + t[[10,19]]); d = Abs[ts-t[[10,20]]] + d;]) In[4]: = Print[“Sweep #”,nr]; Round[N[t,2]] Out[4] = Sweep #2 {{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {10, 4, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {20, 9, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, © 2001 by CRC Press LLC

{30, 13, 5, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {40, 18, 7, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {50, 20, 8, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {17, 11, 5, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, { 6, 5, 3, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, { 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, { 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}} Notice that In[2] initializes the boundary temperatures, nr keeps the count of how many sweeps have been performed, and the function Round is employed in In[4] to round the temperature value to a two-digit integer. When the total temperature differences, d, is limited to eps = 100 degrees, the t values obtained after two sweeps are slightly different from those obtained by the other versions, this is again because Mathematica keeps more significant digits in all computation steps than those in the FORTRAN, QuickBASIC, and MATLAB programs. By changing the eps value from 100 degrees to 1 degree, Mathematica also takes 137 sweeps to converge as in the FORTRAN, QuickBASIC, and MATLAB versions: In[5]: = t = Table[0,{10},{20}]; eps = 1; nr = 0; d = eps + 1; In[6]: = Do[t[[i,1]] = (I–1)*10,{i,2,6}]; In[7]: = Print[“Sweep #”,nr]; Round[N[t,2]] Out[7] = Sweep #137 {{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {10, 8, 7, 5, 4, 3, 3, 2, 2, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0}, {20, 16, 13, 10, 8, 7, 5, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 0, 0, 0}, {30, 24, 19, 15, 12, 9, 8, 6, 5, 4, 3, 3, 2, 2, 1, 1, 1, 1, 0, 0}, {40, 30, 23, 18, 15, 12, 10, 8, 6, 5, 4, 4, 3, 2, 2, 1, 1, 1, 0, 0}, {50, 34, 26, 20, 16, 13, 11, 9, 7, 6, 5, 4, 3, 3, 2, 2, 1, 1, 0, 0}, {35, 30, 25, 21, 17, 15, 12, 10, 8, 7, 6, 5, 4, 3, 3, 2, 2, 1, 1, 0}, {29, 27, 24, 21, 18, 15, 13, 11, 9, 7, 6, 5, 4, 3, 3, 2, 2, 1, 1, 1}, {26, 26, 23, 21, 18, 16, 13, 11, 9, 8, 6, 5, 4, 4, 3, 2, 2, 1, 1, 1}, {26, 25, 23, 21, 18, 16, 13, 11, 9, 8, 7, 5, 4, 4, 3, 2, 2, 2, 1, 1}} The above results all agree with those obtained earlier.

WARPING ANALYSIS

OF A

TWISTED BAR

WITH

NONCIRCULAR CROSS SECTION

As another example of applying the relaxation method for engineering analysis, consider the case of a long bar of uniform rectangular cross section twisted by the two equal torques (T) at its ends. The cross section of the twisted bar becomes warped as shown in Figure 6. If z-axis is assigned to the longitudinal direction of the bar, to find the amount of warping at any point (x,y) of the cross sectioned surface, denoted as W(x,y), the relaxation method can again be employed because © 2001 by CRC Press LLC

If the string is made of a single material, T/m would then be equal to a constant. Analytical solution can be found for this simple case. For the general case that the string may be composed of a number of different materials and the mass m is then a function of the spatial variable x. The more complicated the variation of these properties in x and t, the more likely no analytical solution possible and the problem can only be solved numerically. The finite-difference approximation of Equation 1 can be achieved by applying the central difference for the second derivatives for both with respect to the space variable x and the time variable t. If we consider the displacement u only a finite number of stations, say N, defined with a spatial increment x and using a time increment of t, then specially, for t at ti and x at xj, we can have: ∂2 u u i −1, j − 2 u i, j − u i +1, j = ∂t 2 ( ∆t )2 and ∂2 u u i, j−1 − 2 u i, j + u i, j+1 = ∂x 2 ( ∆x)2 Substituting the above expressions into Equation 1, we obtain: 2

a∆t  u i +1, j = − u i =1, j + 2 u i, j +  u − 2 u i, j + u i, j+1  ∆x  i, j−1

(

)

(7)

Equation 7 is to be applied for j = 2 through j = N–1. Initially for t = t1 = 0, Equation 7 can be applied for i = 2, then ui–1,j≡ u(t1,xj) term is simply f(xj) and can be calculated using Equation 3, and the ui,j–1, ui,j, and ui,j + 1 terms can be calculated using the forward-difference approximations of Equation 4 which are, for k = 2,3,…,N–1 u(t 2 , x k ) = u(t1, x k ) + v0 (x k )∆t = f (x k ) + v 0 (x k )∆t

(8)

Since both f(x) and v0(x) are prescribed, use of Equation 7 enables all u to be calculated at t = t3 and at all in-between stations xj, for j = 2,3,…,N–1. When u values at t = t2 and t = t3 are completely known, Equation 7 can again be utilized to compute u at t = t4 for i = 4 and so forth. Because the string is tightened, we expect the magnitudes of u values to continuously decrease in time. That is, for a specified tolerance, we may demand that the computation be terminated when max. u i, j < ε

j=2~N-1

© 2001 by CRC Press LLC

(9)

In[5]: = t = t + dt; u = u0 + dt*v0; un = Table[0.{j,n + 2}]; nc = 2; In[6]: = (While[exitflag == 0, usum = 0; Do[un[[j + 1]] =-u0[[j + 1]] + c*u[[j]] + 2*(1–c)*u[[j + 1]] + c*u[[j + 2]]; usum = usum + Abs[un[[j + 1]]];,{j,n}]; t = t + dt; If[(usum<eps)||(t>tend), exitflag = 1; Break, u0 = u; u = un;]; If[nc = = np, Print["t = ",N[t,3], " String's Displacements are: "]; Print[N[u,3]]; nc = 1;, nc = nc + 1]]) Out[6]: = t = 0.1 String's Displacements are: {0. 0.167, 0.333, 0.5, 0.667, 0.833, 0.997, 1.14, 1.21, 1.07, 0.795, 0.692, 0.499, 0.359, 0.131, 0.0961, 0} t = 0.2 String's Displacements are: {0. 0.163, 0.315, 0.423, 0.418, 0.24, –0.0295, –0.184, –0.285, –0.515, –0.632, –0.474, –0.749, –0.564, –0.287, –0.168, 0} t = 0.3 String's Displacements are: {0. –0.449, –0.818, –1.01, –1.12, –1.31, –1.47, –1.47, –1.45, –1.22, –0.95, –0.812,–0.691, –0.514, –0.298, –0.193, 0} t = 0.4 String's Displacements are: {0. –0.42, –0.88, –1.26, –1.54, –1.63, –1.49, –1.44, –1.36, –1.17, –0.98, –0.829, –0.7, –0.456, –0.371, –0.146, 0} t = 0.5 String's Displacements are: {0. 0.242, 0.411, 0.285, 0.0344, 0.00989, –0.0857, –0.355, –0.0622, –0.774, –0.804, –0.803, –0.6, –0.528, –0.308, –.177, 0} t = 0.6 String's Displacements are: {0. 0.0971, 0.341, 0.481, 0.695, 0.976, 1.1, 1.01, 0.896, 0.855, 0.68, 0.546, 0.217, 0.0517, –0.0764, –0.0592, 0} t = 0.7 String's Displacements are: {0. 0.183, 0.289, 0.523, 0.706, 0.82, 0.958, 1.11, 1.38, 1.59, 1.82, 1.74, 1.62, 1.44, 1.15, 0.663, 0} t = 0.8 String's Displacements are: {0. 0.207, 0.297, 0.481, 0.724, 0.808, 0.913, 1.17, 1.2, 1.24, 1.1, 0.893, 0.689, 0.567, 0.481, 0.273, 0} t = 0.9 String's Displacements are: {0. 0.164, 0.296, 0.379, 0.495, 0.44, 0.296, 0.0756, –0.104, –0.411, –0.537, –0.51, –0.563, –0.57, –0.537, –0.323, 0} © 2001 by CRC Press LLC

t = 1.0 String's Displacements are: {0. –0.22, –0.459, –0.725, –1.05, –1.15, –1.31, –1.34, –1.28, –1.31 –1.2, –0.888, –0.539, –0.463, –0.53, –0.197, 0} t = 1.1 String's Displacements are: {0. –0.564, –0.992, –1.41, –1.64, –1.81, –1.69, –1.47, –1.25, –1.03, –1.06, –0.878, –0.651, –0.485, –0.361, –0.127, 0} t = 1.2 String's Displacements are: {0. –0.0102, 0.0177, 0.194, 0.133, –0.126, –0.447, –0.714, –0.804, –0.869, –0.789, –0.8, –0.639, –0.479, –0.324, –0.165, 0} t = 1.3 String's Displacements are: {0. 0.134, 0.357, 0.611, 0.813, 0.91, 0.897, 0.944, 0.815, 0.674, 0.465, 0.222, –0.0661, –0.0771, –0.0758, –0.135, 0} t = 1.4 String's Displacements are: {0. 0.227, 0.43, 0.455, 0.57, 0.721, 1.07, 1.3, 1.47, 1.56, 1.67, 1.64, 1.53, 1.38, 1.03, 0.484, 0} t = 1.5 String's Displacements are: {0. 0.207, 0.32, 0.515, 0.561, 0.869, 1.08, 1.12, 1.09, 1.18, 1.27, 1.24, 1.16, 0.882, 0.515, 0.268, 0} t = 1.6 String's Displacements are: {0. 0.122, 0.339, 0.368, 0.499, 0.531, 0.534, 0.281, 0.0675, –0.0461, –0.285, –0.451, –0.579, –0.613, –0.48, –0.205, 0} t = 1.7 String's Displacements are: {0. –0.106, –0.269, –0.467, –0.732, –0.909, –1.19, –1.31, –1.25, –1.2, –1.08, –0.995, –0.837, –0.533, –0.156, 0.00405, 0} t = 1.8 String's Displacements are: {0. –0.585, –1.2, –1.55, –1.65, –1.72, –1.74, –1.58, –1.38, –1.66, –0.903, –0.718, –0.701, –0.542, –0.366, –0.166, 0} t = 1.9 String's Displacements are: {0. 0.0718, –0.0337, –0.188, –0.197, –0.4499, –0.591, –0.741, –0.937, –1.02, –0.919, –0.742, –0.546, –0.481, –0.362, –0.203, 0} t = 2.0 String's Displacements are: {0. 0.0356, 0.543, 0.699, 0.666, 0.816, 0.828, 0.853, 0.708, 0.399, 0.0934, –0.0436, –0.0906, –0.161, –0.192, –0.0974, 0} The results are in complete agreement with those obtained previously. © 2001 by CRC Press LLC

8.5 PROBLEMS PARABPDE 1. Expand the program ParabPDE to print out the computed temperature distribution along the entire rod only when the temperature at an interactively specified location reaches an interactively entered value. The time required to reach this condition should be printed as shown below. Call this new program ParaPDE1. Enter the x value at which a desired temperature is to be specified: 6.0 Enter the temperature to be reached, in °F: 50 It takes X.XXXXXE + 00 seconds. Notice that a format of E13.5 is to be used to print out the time. 2. Consider the transient heat-conduction problem where k/cρ = 0.00104 ft2/sec, t = 1 sec, x = 1”, and at the left end of the rod, x = 0, the temperature u is equal to 50°F initially but equal to 0°F for t>0. Compute the temperatures at x = 1”, 2”, and 3” when t = 1, 2, and 3 seconds. 3. Use the same data as in the program ParabPDE, but modify the program for the case when the right end is not insulated but is maintained at 0°F until t = 100 seconds and then is heated at 100°F afterwards. Print out the times required for the station at the mid-length reaches to 10°F, 20°F, at 10°F increments until 100°F. 4. Study the effect of changing the value of DT in the program ParabPDE on tfinal, the time required for reaching the uniform distribution of 100°F throughout the rod. Tabulate DT vs. tfinal. 5. Change the steady-state temperature from 100°F to 75°F and generate a plot similar to that shown in Figure 2. 6. For the rod shown in Figure 1, one-half of the surface insulation, for x/L = 0.4 to x/L = 0.7, is to be removed and the temperature for that portion of the rod is to be maintained at 70°F. Modify the program ParabPDE is accommodate for the computation needed to determine how long it takes to reach a stable temperature distribution. 7. Apply MATLAB and Mathematica (using ParaPDE.m) to solve the transient heat-conduction problem by decreasing x from 1 to 0.5.

RELAXATN 1. For the problem treated by the program Relaxatn except that the temperature is equal to 100°F along the top boundary, perform the relaxation by upgrading T starting from the top boundary instead of from the bottom boundary to expedite the convergence. Modify the program Relaxatn to perform this new sweeping process. 2. Round the right lower corner of the square plate with a radius equal to 5 and consider this corner as an insulated boundary. Modify the program © 2001 by CRC Press LLC

the program Gauss to find these MxN W’s. Compare the resulting W distribution with those obtained by the relaxation method shown in Figures 5 and 6 for a = 30 and b = 20. 12. Same as Problem 11 except for the case a = b = 20 and for comparing with Figure 7. 13. Solve the warping problem by Mathematica.

WAVEPDE 1. For the string problem analyzed in the Sample Application, modify the program slightly so that the times required for the string to have the magnitudes of its maximum displacements reduced to 0.8, 0.6, 0.4, and 0.2, and the corresponding deflected shapes can be printed. 2. Rearrange the subprogram FUNCTION F in the program WavePDE to consider the case of an initial, upward lifting the mid-third (8≤x≤16 cm) of the string by 1 cm. Rerun the program using the same input data as in the Sample Application. 3. Consider a string which is composed of two different materials even though it is subjected to a uniform tension T so that the left and right onethirds (i.e., , 0≤x≤8 cm and 24≤x≤32 cm, respectively) of the string has a wave velocity a = 80 cm/sec while its mid-third (i.e., 8≤x≤16 cm) has a wave velocity a = 90 cm/sec. Modify and then rerun program WavePDE using the other input same as in the Sample Application. 4. A tightened string of length L equal to 1 ft is lifted as shown in 15 and is released with a velocity distribution v = y(t = 0,x)/t = 2sin x/L in ft/sec. If the constant T/m appearing in Equation 2 is equal to 8,100 ft2/sec2, use a time increment t = 0.0005 sec and a space increment x = 0.1L and apply Equation 7 to find the y values at t = 0.001 sec and for the stations x2 and x3. 5. In approximating Equation 1 by finite differences, we may keep the same approach for 2u/t2 as in deriving Equation 7 but to apply the second central-difference formula for 2u/t2 not at t = ti but at t = ti + 1. The resulting equation, for C = (at/x)2, is: −Cu i +1, j−1 + (2 + C)u i +1, j − Cu i +1, j+1 = − u i −1, j + 2 u i, j Derive a matrix equation for solving the unknowns uj for j = 1,2,…,N–1, at t = ti + 1. Note that the boundary conditions are ui + 1,0 = ui + 1,N = 0. Write a program WavePDE.G which uses the Gaussian Elimination method to solve this matrix equation and run it for the Sample problem to compare the results. 6. Change the MATLAB m file WavePDE to solve Problem 4. 7. Apply Mathematica to solve Problem 4.

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8.6 REFERENCES 1. C. R. Wylie, Jr., Advanced Engineering Mathematics, Chapter 9, Second Edition, McGraw-Hill, New York, 1960. 2. J. P. Holman, Heat Transfer, McGraw-Hill, New York, 1963. 3. S. Timoshenko and J. N. Goodier, Theory of Elasticity, Chapter 11, Second Edition, McGraw-Hill, New York, 1951.

© 2001 by CRC Press LLC